Further Pure | Chapter 8: Differential Equations

Curriculum Objectives

• recall the meaning of the terms ‘complementary function’ and ‘particular integral’ in the context of linear differential equations, and recall that the general solution is the sum of the complementary function and a particular integral;
• find the complementary function for a second order linear differential equation with constant coefficients;
• recall the form of, and find, a particular integral for a second order linear differential equation in the cases where a polynomial or ebx or a cos px + b sin px is a suitable form, and in other simple cases find the appropriate coefficient(s) given a suitable form of particular integral;
• use a substitution to reduce a given differential equation to a second order linear equation with constant coefficients;

Note: This chapter is written rather hypocritically. I am a firm believer of understanding over rote learning, yet in this chapter, often you will be required to learn the procedure, without being given an explanation, as to their need. This chapter requires some learning and will not over-complicate things by digressing into the proofs or reasoning behind theorems.

Introduction to the Differential Equation

You must already be acquainted with the general idea of a differential equation. Yet a concise introduction is presented here.

The fundamental definition of a differential equation is that it is an equation that involves differential coefficients. Consider the following:

$\begin{array}{l} i)\;\frac{{dy}}{{dx}} = 2x\\ ii)\,y\frac{{{d^2}y}}{{d{x^2}}} + x = 0\\ iii)y = x \cdot \frac{{dy}}{{dx}} + \frac{5}{{\frac{{dy}}{{dx}}}} \end{array}$

When we talk about solving differential equations, we wish to find a function of the independent variable that it satisfies the original equation. If you were given the differential equation:

$\frac{{dy}}{{dx}} = 2$

You can easily deduce that the solution to this differential equation would be such a function, y, that when differentiated yields, 2. The solution of course is an entire "family" defined by: $y = 2x + c$

We shall introduce here the Second Order Linear Differential Equation, with constant coefficients ( the co-efficient themselves can be functions ), for it's assumed you have already met equations of the first order in the Mathematics course. The order of a differential equation tells us the highest order of derivative that is present in the equation. A second order differential equation will, then, always involve a function twice differentiated. The general differential equation of this sort is of the form:

$a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$

Here, a cannot be equal to zero, for then it will not be a second order differential equation. The other constants do not function under any such condition.

Obviously we will begin with the easier case, and only take differential equations of the form

$a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0$

Such equations are called homogeneous equations.

Ex. Solve the differential equation:

$16y - \frac{{{d^2}y}}{{dx}} = 0$

Sol.

By inspection and some ingenuity alone I can deduce that the following will be the solution to the differential equation, as we require a function, such that differentiated twice,, equals 16 times the original function:

$y = {e^{4x}};{e^{ - 4x}}$

These alone aren't the only possible solutions. All of the expressions of the form $y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}}$ will be a solution to this differential equation. You make check this by twice differentiating the expression and substituting for both y and y''. 

This is called the Principle of Superposition. If you know ${y_1},\,{y_2}$ to be solutions of a linear, homogeneous ( these are the equations that we are considering, equations that are equated with zero as opposed to some function of x ), differential equation, then $y = {c_1}{y_1} + {c_2}{y_2}$ is also a solution. This fact is the very definition of a linear differential equation. The solutions of a linear differential equation form a linear space ( you shall meet this in the next chapter ).

The presence of constants in the solution suggests that it can model a variety of situations. For a specific case, however, we will have to put in the values of the constants. The values are found from boundary conditions. For eg, if we are given the following boundary conditions for this question:

$y = {c_1}{e^{4x}} + {c_2}{e^{ - 4x}}$

This solution is called the complementary function.

$\begin{array}{l} y = 2,\quad when\,x = 0\\ \frac{{dy}}{{dx}} = - 1,\quad when\,x = 0\\ {c_1} + {c_2} = 2\\ \frac{{dy}}{{dx}} = 4{c_1}{e^{4x}} - 4{c_2}{e^{ - 4x}}\\ 4{c_1} - 4{c_2} = - 1\\ Solving\,this\,system\,of\,equations,\,we\,have:\\ {c_1} = \frac{7}{8};\,{c_2} = \frac{9}{8}\\ y = \frac{7}{8}{e^{4x}} + \frac{9}{8}{e^{ - 4x}} \end{array}$


In the original example we were able to deduce the answer by inspection, this doesn't occur very often in the examination. So let us then generalise our method.

Let me assure you that the solution of these equations is of the form $y = {e^{mx}}$ ( This particular function is our best guess at the answer) . The question is then, what is m?

We employ the following method:

$\begin{array}{l} a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0\\ y = {e^{mx}}\\ \frac{{dy}}{{dx}} = m{e^{mx}}\\ \frac{{{d^2}y}}{{d{x^2}}} = {m^2}{e^{mx}}\\ a\left( {{m^2}{e^{mx}}} \right) + b\left( {m{e^{mx}}} \right) + c{e^{mx}} = 0\\ {e^{mx}}\left( {a{m^2} + bm + c} \right) = 0\\ {e^{mx}} \ne 0\quad \therefore a{m^2} + bm + c = 0 \end{array}$

The roots of this quadratic are then our values of m. This quadratic is of great importance to the solution to second order linear ODE, and therefore warrants its own name. Its called the characteristic equation.

There are three possibilities then for m. Either it has two completely different roots as its values, a repeated root as its only value, or a complex root and its conjugate as its value. These cases differ in treatment, therefore will be discussed separately.

Real roots

If we get two real roots to our characteristic equation ($m_1$ and $m_2$), then the solution will be of the form:

$y = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}$

Ex. Solve

$\frac{{{d^2}y}}{{d{x^2}}} + 11\frac{{dy}}{{dx}} + 24y = 0$

Sol.

The Characteristic equation is then:

$\begin{array}{l} {m^2} + 11m + 24 = 0\\ \left( {m + 3} \right)\left( {m + 8} \right) = 0 \end{array}$

The general solution is then:

$y = {c_1}{e^{ - 3x}} + {c_2}{e^{ - 8x}}$

Repeated Roots

For repeated roots, the general solution is of the form:

$y = \left( {{c_1} + {c_2}x} \right){e^{rx}}$

where r is the repeated root of the equation.

Ex.

$\frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 4y = 0;\quad {y_{x = 0}} = 12;\quad {\frac{{dy}}{{dx}}_{x = 0}} =  - 3$

Sol.

$\begin{array}{l} \frac{{{d^2}y}}{{d{x^2}}} - 4\frac{{dy}}{{dx}} + 4y = 0;\quad {y_{x = 0}} = 12;\quad {\frac{{dy}}{{dx}}_{x = 0}} = - 3\\ {r^2} - 4r + 4 = 0\\ {\left( {r - 2} \right)^2} = 0\\ r = 2\\ y = {c_1}{e^{2x}} + {c_2}x{e^{2x}}\\ \frac{{dy}}{{dx}} = 2{c_1}{e^{2x}} + {c_2}{e^{2x}} + 2{c_2}x{e^{2x}}\\ {c_1} = 12;{c_2} = - 27\\ y = 12{e^{2x}} - 27x{e^{2x}} \end{array}$

Complex Roots

For complex roots the solution is of the form:

$y = {e^{{r_1}x}}\left( {{c_1}\cos \left( {{r_2}x} \right) + {c_2}\sin \left( {{r_2}x} \right)} \right)$

Where the complex roots of the characteristic equation are: ${r_1} \pm \iota {r_2}$


Ex. Solve the equation $y'' + 6y' + 9y = 0$ given that $\frac{{dy}}{{dx}} = 6;\;y = 0;\;and\;x = 0$

Sol.

Solving the characteristic equation I find:

$\begin{array}{l} {m^2} + 6m + 9 = 0\\ m = 1 \pm \iota \sqrt 2 \end{array}$

The general solution is then of the form:

$y = {e^x}\left( {A\cos \left( {\sqrt 2  \cdot x} \right) + B\sin \left( {\sqrt 2  \cdot x} \right)} \right)$

Differentiating this equation, gives me:

$y' = {e^x}\left( {A\cos \left( {\sqrt 2  \cdot x} \right) + B\sin \left( {\sqrt 2  \cdot x} \right) + \sqrt 2 \left( {B\cos \left( {\sqrt 2  \cdot x} \right) - A\sin \left( {\sqrt 2  \cdot x} \right)} \right)} \right)$

Using the boundary conditions to create a system of equations and then solving it, I get the following final result:

$y = 3\sqrt 2 \sin \left( {\sqrt 2  \cdot x} \right)$

The reader is, as always, encouraged to solve this on his own and convince his self of this answer.

Differential Equations of the form $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$

For such equations the solution is found by working out the complementary function and particular integral. The general solution is then the sum of these two.

The complementary function of such equations, $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = f(x)$, is found by isolating the left hand side, i.e. $a\frac{{{d^2}y}}{{d{x^2}}} + b\frac{{dy}}{{dx}} + cy = 0$ and then solving it as we have done previously. The solution we shall call the complementary function

The particular integral is found using the entire differential equation, but depends heavily upon the function on the right side of the equation. The following explains the method: 

Consider the following the examples

Ex. Solve $y'' + 3y' - 4y = 8$

The complementary function, is found by solving the characteristic equation, The reader will find that the complementary function is:

$y = {c_1}{e^{ - 4x}} + {c_2}{e^x}$

For the particular integral we see that the function on the right hand side is a polynomial of the degree zero, which is to say that it is a constant. We assume that particular integral will also be of the form, and hence let $y = k$. Plugging this into the equation:

$\begin{array}{l} y = k,\,y' = 0,\,y'' = 0\\ y'' + 3y' - 4y = 8\\ - 4k = 8\\ k = - 2\\ y = - 2 \end{array}$

The particular integral is then, $y = -2$. The general solution is then:

$y = {c_1}{e^{ - 4x}} + {c_2}{e^x} - 2$

If boundary conditions were given in the question, it would be at this point, that you plug in that information to work out the constants

Ex. Solve $y'' + 3y' - 4y = 2 + 8{x^2}$

The complementary function is the same. For the particular integral, we note that there is a quadratic on the right side. Hence we use a quadratic of our own, to work out the particular integral. We let $y = a{x^2} + bx + c$

$\begin{array}{l} y' = 2ax + b\\ y'' = 2a \end{array}$
$\begin{array}{l} y'' + 3y' - 4y = 8{x^2} + 2\\ 2a + 3\left( {2ax + b} \right) - 4\left( {a{x^2} + bx + c} \right) = 8{x^2} + 2\\ 2a + 6ax + 3b - 4a{x^2} - 4bx - 4c = 8{x^2} + 2\\ \therefore - 4a{x^2} = 8{x^2} \Rightarrow a = - 2\\ \left( {6a - 4b} \right)x = 0x \Rightarrow b = - 3\\ 2a + 3b - 4c = 2 \Rightarrow c = - 4 \end{array}$

The general solution is of the form:

$y = {c_1}{e^{ - 4x}} + {c_2}{e^x} - 2{x^2} - 3x - 4$

So, generally, if there is a polynomial on the right hand side of order n, you must use the following to find out your particular integral, $y = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_n}{x^n}$.

For the right hand side, we still have two possibilities to exhaust. Consider further the examples:

$y'' + 3y' + 2y = 3{e^{2x}}$

The complementary function is left to the reader to work out. We shall focus on the particular integral. We shall here use as our particular integral, the general form: $y = k{e^{2x}}$.

$y = k{e^{2x}};\,y' = 2k{e^{2x}};\,y'' = 4k{e^{2x}}$

As before, we put in our guess at the particular integral into the equation to give:

$\begin{array}{l} 4k{e^{2x}} + 3\left( {2k{e^{2x}}} \right) + 2\left( {k{e^{2x}}} \right) = 3{e^{2x}}\\ 12k = 3 \Rightarrow k = {\textstyle{1 \over 4}} \end{array}$

The particular integral is therefore $y = {\textstyle{1 \over 4}}{e^{2x}}$

Lastly consider:

$y'' + 3y' + 2y = \cos \left( {2x} \right)$

Here we find cosine on the right hand side. If we had sine on the right hand side, or a combination of the two ratios, our guess would be the same, which is: $a\cos \left( {2x} \right) + b\sin \left( {2x} \right)$


$\begin{array}{l} y = a\cos \left( {2x} \right) + b\sin \left( {2x} \right)\\ y' = - 2a\sin \left( {2x} \right) + 2b\cos \left( {2x} \right)\\ y'' = - 4a\cos \left( {2x} \right) - 4\sin \left( {2x} \right) \end{array}$

We put this into the initial equation:

$ - 4a\cos \left( {2x} \right) - 4\sin \left( {2x} \right) + 3\left( { - 2a\sin \left( {2x} \right) + 2b\cos \left( {2x} \right)} \right) + 2\left( {a\cos \left( {2x} \right) + b\sin \left( {2x} \right)} \right) = \cos \left( {2x} \right)$

Comparing terms in, sin(2x) and cos(2x), on either sides yields:

$\begin{array}{l} - 4a + 6b + 2a = 1\\ - 4b - 6a + 2b = 0 \end{array}$

This system of equations can be solved to give:

$a =  - {\textstyle{1 \over {20}}};\,b = {\textstyle{3 \over {20}}}$

The particular integral is then: $y =  - {\textstyle{1 \over {20}}}\cos \left( {2x} \right) + {\textstyle{3 \over {20}}}\sin \left( {2x} \right)$

The following table summarises these important rules; you should commit it to memory.

It might happen that the general expression you choose to derive the particular integral with, may be excepting the constants, is equivalent to the complementary function you have already derived. In this case you must amend the general expression you were going to use, by multiplying the general expression by the independent variable. Consider the example:

Ex. Solve $y'' - 3y' = 6$

It is easily seen that the complementary function is:

$y = A + B{e^{3x}}$

For the particular integral, I take the general expression, y = k. It should have worked, yet putting this into the initial equation, one sees:

$\begin{array}{l} y = k;\,y' = 0;\,y'' = 0\\ 0 = 6 \end{array}$

This happens because our particular integral is already present in the complementary function. A, a constant, is present in the CF, and we have chosen a constant for our PI. Here we multiply our PI by the independent variable, in this case (x). Our particular integral is then $ y = kx $. It is left to the reader to ascertain that the PI, is $ y = - 2x $

If the function on, right hand side been assume, $7{e^{3x}}$, we could not have used as our particular integral, $k{e^{3x}}$, rather we would have used: $kx{e^{3x}}$. Had this also been present in the CF, we would have used: $k{x^2}{e^{3x}}$

Using a substitution to reduce a given differential equation to a second order linear equation with constant coefficients

In your examination, you may be given a differential equation involving, say y and x as variables, that can be reduced to a second order linear equation with constant coefficient of say, z and x, provided a relationship between the variables is given. In most cases, you will be given an intimidating differential equation, the equation it needs to be reduced to, and the substitution. Consider the following example.

Ex. Solve $\frac{1}{x}\frac{{{d^2}y}}{{d{x^2}}} + \left( {\frac{6}{x} - \frac{2}{{{x^2}}}} \right)\frac{{dy}}{{dx}} + \left( {\frac{9}{x} - \frac{6}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)y = 169\sin \left( {2x} \right)$ given that $y = xz$

Sol:

We note that:

$\begin{array}{l} \frac{{dy}}{{dx}} = z + x\left( {\frac{{dz}}{{dx}}} \right)\\ \frac{{{d^2}y}}{{d{x^2}}} = \frac{{dz}}{{dx}} + \frac{{dz}}{{dx}} + x\left( {\frac{{{d^2}z}}{{d{x^2}}}} \right) \end{array}$

Substituting these into the original equation:

$\begin{array}{l} \frac{1}{x}\left( {2\left( {\frac{{dz}}{{dx}}} \right) + x\left( {\frac{{{d^2}z}}{{dx}}} \right)} \right) + \left( {\frac{6}{x} - \frac{2}{{{x^2}}}} \right)\left( {z + x\left( {\frac{{dz}}{{dx}}} \right)} \right) + \left( {\frac{9}{x} - \frac{6}{{{x^2}}} + \frac{2}{{{x^3}}}} \right)\left( {xz} \right) = 169\sin \left( {2x} \right)\\ \frac{2}{x}\left( {\frac{{dz}}{{dx}}} \right) + \frac{{{d^2}z}}{{d{x^2}}} + \frac{6}{x}\left( z \right) + 6\left( {\frac{{dz}}{{dx}}} \right) - \frac{{2z}}{{{x^2}}} - \frac{2}{x}\left( {\frac{{dz}}{{dx}}} \right) + 9z - \frac{6}{x}z + \frac{2}{{{x^2}}}z = 169\sin \left( {2x} \right)\\ \frac{{{d^2}z}}{{d{x^2}}} + 6\frac{{dz}}{{dx}} + 9z = 169\sin \left( {2x} \right) \end{array}$

Now you may easily solve the differential equation using the method already taught. Alternatively you could have rearranged it thusly: z = y/x. Differentiated it, and compared with the original equation.

Last I shall attempt a question from your past papers.

Ex. (May/June 2005, Question No. 11 OR)

It is given that:

$\frac{{{d^2}y}}{{d{x^2}}} + \left( {2a - 1} \right)\frac{{dy}}{{dx}} + a\left( {a - 1} \right)y = 2a - 1 + a\left( {a - 1} \right)x$

Find y in the terms of a nand x, given that y and y' are zero, when x is zero; a is a constant.

Sol.

The characteristic equation will be made thusly:

${m^2} + \left( {2a - 1} \right)m + {a^2} - a = 0$

M is then:

$\begin{array}{l} m = \frac{{\left( {1 - 2a} \right) \pm \sqrt {{{\left( {2a - 1} \right)}^2} - 4\left( 1 \right)\left( {{a^2} - a} \right)} }}{2}\\ m = \frac{{\left( {1 - 2a} \right) \pm \sqrt {\left( {4{a^2} - 4a + 1} \right) - 4{a^2} + 4a} }}{2} = \frac{{\left( {1 - 2a} \right) \pm 1}}{2}\\ m = 1 - a;\, - a \end{array}$

CF is then:

$y = {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}}$

For the particular integral, we take $y = {c_3}x + {c_4}$

$y = {c_3}x + {c_4};\,y' = {c_3};\,y'' = 0$

We substitute this into the initial equation:

$\left( {2a - 1} \right)\left( {{c_3}} \right) + \left( {{a^2} - a} \right)\left( {{c_3}x + {c_4}} \right) = 2a - 1 + \left( {{a^2} - a} \right)x$

$\begin{array}{l} \left( {{a^2} - a} \right){c_3}x = \left( {{a^2} - a} \right)x \Rightarrow {c_3} = 1\\ \left( {2a - 1} \right){c_3} + \left( {{a^2} - a} \right){c_4} = 2a - 1 \Rightarrow {c_4} = 0 \end{array}$

The particular integral then becomes: $y = x$

The general solution to the equation becomes:

$y = x + {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}}$

$\frac{{dy}}{{dx}} = \left( {1 - a} \right){c_1}{e^{\left( {1 - a} \right)x}} - a{c_2}{e^{ - ax}}$

Using boundary conditions, we have:


$\begin{array}{l} y = {c_1}{e^{\left( {1 - a} \right)x}} + {c_2}{e^{ - ax}} + x \Rightarrow 0 = {c_1} + {c_2} + 0\\ \frac{{dy}}{{dx}} = \left( {1 - a} \right){c_1}{e^{\left( {1 - a} \right)x}} - a{c_2}{e^{ - ax}} + 1 \Rightarrow 0 = \left( {1 - a} \right){c_1} - a{c_2} + 1\\ {c_1} = - {c_2}\\ 0 = {c_1} - a{c_1} - a{c_2} + 1 \Rightarrow {c_1} = - 1\\ y = x + {e^{ - ax}} - {e^{\left( {1 - a} \right)x}} \end{array}$

Ex. Hence show that $y \approx x$ as $x \to \infty $

Sol.

We note:

$y = x + {e^{ - ax}} - {e^{\left( {1 - a} \right)x}}$

If $a > 1$, both the terms - -a and (1 - a) - will be negative. As $x \to \infty$ the terms - ${e^{ - ax}}$ and ${e^{\left( {1 - a} \right)x}}$ - will approach zero, and we will be left with:

$y \approx x$

Further Pure | Chapter 7: Vectors

Curriculum Outcome:

• use the equation of a plane in any of the forms ax + by + cz = d or r.n = p or r = a + λb + µc, and convert equations of planes from one form to another as necessary in solving problems;
• recall that the vector product a × b of two vectors can be expressed either as |a| |b| sin θ nˆ , where nˆ is a unit vector, or in component form as (a2b3 – a3b2) i + (a3b1 – a1b3) j + (a1b2 – a2b1) k;
• use equations of lines and planes, together with scalar and vector products where appropriate, to solve problems concerning distances, angles and intersections, including: determining whether a line lies in a plane, is parallel to a plane or intersects a plane, and finding the point of intersection of a line and a plane when it exists, finding the perpendicular distance from a point to a plane, finding the angle between a line and a plane, and the angle between two planes, finding an equation for the line of intersection of two planes, calculating the shortest distance between two skew lines. finding an equation for the common perpendicular to two skew lines

We shall not tarry on topics you should have learnt in the A Levels Mathematics Course. The majority of the questions on the topic have been simply about the applications of different formulae, that too, mostly straightforward. This is in my reckoning the easiest chapter in your syllabus.

Equation of a Plane

If we know a vector perpendicular to the plane, n, we wish to find an equation to, and further a point on the plane, a, and the general point r, the equation is:

$r \cdot \vec n = a \cdot \vec n$

Or if we use matrices to represent the vectors:

$\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{n_1}}\\ {{n_2}}\\ {{n_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a\\ b\\ c \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{n_1}}\\ {{n_2}}\\ {{n_3}} \end{array}} \right)$

An alternative method of finding the vector equation would require non-collinear points (A, B, C) on the plane. If we have the position vectors of such three points, i.e.

 $OA = a;\quad OB = b;\quad OC = c$

Then the equation of the plane is:

$r = \vec a + m\left( {\vec b - \vec a} \right) + n\left( {\vec c - \vec a} \right)$

Ex. Find the vector equation of the plane containing the points $a = j + k;\quad b = i + j;\quad c = i + k$ and convert it to the cartesian form.

Sol. Given in the question is:

$a = j + k;\quad b = i + j;\quad c = i + k$

The plane then becomes:

$\begin{array}{l} r = j + k + m\left( {\left( {i + j} \right) - \left( {j + k} \right)} \right) + n\left( {\left( {i + k} \right) - \left( {j + k} \right)} \right)\\ r = j + k + m\left( {i - k} \right) + n\left( {i - j} \right) \end{array}$

Comparing it with the cartesian form, which is:

$r = xi + yj + zk$

I get:

$x = m + n;\quad y = 1 - n;\quad z = 1 - m$

Substituting in the values of m and n, into the equation involving x, I get:

$\begin{array}{l} x = \left( {1 - z} \right) + \left( {1 - y} \right)\\ x + y + z = 2 \end{array}$

Vector Product (Cross Product)

In the A Levels Mathematics Course, you must have met the scalar product or the dot product, which is one manner of multiplying two vector. The multiplication of vectors actually involves a couple, one moiety of which is the Vector Product. Perhaps so named, for it produces a vector as a result, in contradiction to the scalar product which produces a scalar quantity.

We define the vector product of the above two vectors, as:

$\vec a \times \vec b = \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \hat n$

The right hand side has the product of the moduli of the vectors, the sine of the acute angle between the vectors, and a unit vector in the direction determined by the right hand rule. For $a \times b$ one curls the fingers from a to b, and the direction of the unit vector is in the direction of the thumb. For $b \times a$ we have therefore:

$\vec b \times \vec a = \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \left( { - \hat n} \right) =  - \left| a \right|\left| b \right|\sin \theta {\kern 1pt} \hat n$

We have that:

$\left( {\vec b \times \vec a} \right) + \left( {\vec a \times \vec b} \right) = 0$

It may also be surmised that the vector product is not, in general, commutative, i.e.

$\left( {\vec b \times \vec a} \right) \ne \left( {\vec a \times \vec b} \right)$

Also note:

$a \times a = 0$

A simple example to give you feel of the matter.

Ex. Find the cross product of the vectors $a = 2i;\quad b = 5j$.

Sol.

As the vectors lie on different axes the angles between them is 90 degrees. the relevant unit vector is the z axis. The Vector Product is then:

$a \times b = \left( 2 \right)\left( 5 \right)\sin \left( {90} \right)\left( k \right) = 10k$

Now let us detail the method you must have met in before. If the vectors are given in the matrix form, or can be brought into it, the cross product is defined for two vectors as:

$\left( {\begin{array}{*{20}{c}} {{a_1}}\\ {{a_2}}\\ {{a_3}} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}}\\ {{b_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{a_2}{b_3} - {a_3}{b_2}}\\ {{a_3}{b_1} - {a_1}{b_3}}\\ {{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right)$

You can either learn this, as it is, or you may learn the determinant approach. For the two vectors considered above, we produce a matrix of the form below and find the determinant of it. If  matrices and determinants are notions, oblivious to you, you may completely ignore this part and move on. It will be revisited in great detail when we begin the chapter linear algebra.

$\left( {\begin{array}{*{20}{c}} i&j&k\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right)$

The determinant of this matrix is:

$\left| {\begin{array}{*{20}{c}} i&j&k\\ {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right| = i\left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_3}}\\ {{b_2}}&{{b_3}} \end{array}} \right| - j\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}\\ {{b_1}}&{{b_3}} \end{array}} \right| + k\left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}\\ {{b_1}}&{{b_2}} \end{array}} \right| = i\left( {{a_2}{b_3} - {a_3}{b_2}} \right) - j\left( {{a_3}{b_1} - {a_1}{b_3}} \right) + k\left( {{a_1}{b_2} - {a_2}{b_1}} \right)$

where i, j, k, are unit vectors in the direction of the axes x, y, z.

Use of Equation in different problems

What follows are formulae and strategies you may require for question on vectors.

Distance of Point from Plane

The distance between the point A (${a_1},\,{a_2},\,{a_3}$) and the plane defined by  $r \cdot n = d$ is given by:

$l = \frac{{\left| {d - b \cdot n} \right|}}{{\left| n \right|}}$

As a special case, the distance from the origin to a plane is:

${l_O} = \frac{{\left| d \right|}}{{\left| n \right|}}$

The sign of d tells us whether the plane lies above or below the origin.

Distance of a point from a line

There are many ways one may be work out the shortest distance between a point and a line. You must already be aware of at least one from your work in the Mathematics Course. What follows is a method that involves the vector product.

Consider the following case. The line is such that it pass through A, and has the direction vector v, i.e. $r = A + \lambda V$. The point in consideration is S. If we consider the vectors AS, and V, then:

$\begin{array}{*{20}{l}} {\overrightarrow {AS} \times \overrightarrow V = \left| {AS} \right|\left| V \right|\sin \theta \hat n}\\ {\left| {AS \times V} \right| = \left| {\left| {AS} \right|\left| V \right|\sin \theta \hat n} \right|}\\ {\left| {AS \times V} \right| = \left| {AS} \right|\left| V \right|\sin \theta }\\ {\frac{{\left| {AS \times V} \right|}}{{\left| V \right|}} = \left| {AS} \right|\sin \theta }\\ {\frac{{\left| {AS \times V} \right|}}{{\left| V \right|}} = l} \end{array}$

For the above method we realise that the modulus of a unit vector is 1 and that $\left| {AS} \right|\sin \theta $ is the length that is to be found.

Determining whether a line lies in a plane

If we have a line, $ r = \vec a + t\vec b$, and a plane, $ax + by + cz = d$, the method to determine whether the line lies on the plane is given below.

Ex.

Given that:

$\begin{array}{l} r = \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ x + y + z = 2 \end{array}$

You can either substitute the co-ordinates of a general point directly into the plane, and hence prove that any general point on the plane also satisfies the plane equation:

$\begin{array}{l} \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0 + t}\\ {1 - t}\\ 1 \end{array}} \right)\\ x + y + z = 2\\ \left( {0 + t} \right) + \left( {1 - t} \right) + 1 = 2\\ 2 = 2 \end{array}$

Alternatively you could prove that some point on the line lies on the plane as well, and that the direction vector is parallel to the plane:

$\begin{array}{l} r = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ x + y + z = 2\\ \left( 0 \right) + \left( 1 \right) + 1 = 2\\ 2 = 2\\ n = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right);\;\vec b = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right)\\ n \cdot b = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 1 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 0 \end{array}} \right) = 1 - 1 + 0 = 0 \end{array}$

The dot product between the normal to the plane and the direction vector of the line is taken to prove that the direction vector, and in turn the line, is parallel to the plane. As is clear, this all follows from what you have already studied and I am teaching nothing new. Following this are also strategies and formulae.

Determine whether the line is parallel or perpendicular to the plane 

To determine whether it is parallel you follow the second method outlined in the previous section, though you will find that the point may not lie in the plane. You could also take a vector lying in the plane, and find the cross product of that vector and the direction vector of the line. If the resultant vector is a multiple of the normal of the plane, the line is parallel to the plane.

To prove that a line is perpendicular to a plane, one finds a vector lying in the plane and take its dot product with the direction vector of the line. If the dot product is zero, the line is parallel.

Angle between line and plane

Let there be a line L and a plane $\pi$:

$\begin{array}{l} L = a + t\vec b\\ \pi :r \cdot n = d \end{array}$

where the second line must be taken to mean, that the plane is defined by the following equation. Also note that the choice of $\pi$ is purely arbitrary and it has nothing to do with the number defined as the ratio of a circle's circumference to its diameter. Regardless, let us say that $\theta$ is the acute angle between the line and the plane. Then:

$\theta  = {\sin ^{ - 1}}\left( {\frac{{n \cdot d}}{{\left| n \right|\left| d \right|}}} \right)$

Angle between two planes

The angle between two planes, is the angle between the two normals to the planes. The angle is therefore:

$\theta  = {\cos ^{ - 1}}\left( {\left| {\frac{{{n_1} \cdot {n_2}}}{{\left| {{n_1}} \right|\left| {{n_2}} \right|}}} \right|} \right)$ 

where ${n_i}$ are the normals to the planes

Line of Intersection of two planes

The determination of the lines of intersection is illustrated through the following example:

Ex. Determine the line of intersection of the planes

$\begin{array}{l} r \cdot \left( {i + j + k} \right) = 2\\ r \cdot \left( {i + 2j + 3k} \right) = 3 \end{array}$

To begin with we write down the cartesian forms:

$\begin{array}{l} x + y + z = 2\\ x + 2y + 3z = 3 \end{array}$

We let $x = \lambda$ ( or you could let y or z be any variable of your choosing )

$\begin{array}{l} y + z = 2 - \lambda \\ 2y + 3z = 3 - \lambda \end{array}$
$\begin{array}{l} 2y + 3z - 2\left( {y + z} \right) = 3 - \lambda - 4 + 2\lambda \Rightarrow z = \lambda - 1\\ 2y + 3z - 3\left( {y + z} \right) = 3 - \lambda - 6 + 3\lambda \Rightarrow y = 3 - 2\lambda \end{array}$

Using this information one can produce the following result:

$\left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} \lambda \\ { - 1 + \lambda }\\ {3 - 2\lambda } \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ { - 1}\\ 3 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 2} \end{array}} \right)$

Shortest distance between two skew lines

Given two lines:

$\begin{array}{l} {r_1} = {a_1} + \lambda {b_1}\\ {r_2} = {a_2} + \lambda {b_2} \end{array}$

The shortest distance is given by:

$l = \frac{{\left| {\left( {{a_1} - {a_2}} \right) \cdot {b_1} \times {b_2}} \right|}}{{\left| {{b_1} \times {b_2}} \right|}}$

The expression in the numerator is called a scalar triple product. You are to compute the vector product first and the scalar product second, but this much should be obvious as it wouldn't work the other way around.

Finding an equation for the common perpendicular to two skew lines

Allow me to explain with an example:

Ex. Find the points of intersection of a line that is mutually perpendicular to the skew lines $L_1$ and $L_2$, and hence find the equation this line.

$\begin{array}{l} {L_1} = \left( {\begin{array}{*{20}{c}} 3\\ { - 1}\\ 2 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right)\\ {L_2} = \left( {\begin{array}{*{20}{c}} 0\\ 5\\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) \end{array}$

Sol.

The Line mutually perpendicular to both lines, has its direction vector as the cross product of the direction vectors of $L_1$ and $L_2$. Hence:

$\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ { - 3} \end{array}} \right)$

General Points of the two lines are:

$\begin{array}{l} {r_1} = \left( {\begin{array}{*{20}{c}} {3 + s}\\ {s - 1}\\ 2 \end{array}} \right)\\ {r_2} = \left( {\begin{array}{*{20}{c}} t\\ {5 - 2t}\\ {2 + t} \end{array}} \right) \end{array}$

The vector that joins these two lines is then:

$\vec v = \left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right)$

This vector should be now perpendicular to the direction vectors of the initial lines.

$\left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0 \end{array}} \right) = 2s + t - 3 = 0$
$\left( {\begin{array}{*{20}{c}} {3 + s - t}\\ {s + 2t - 6}\\ { - t} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 1 \end{array}} \right) = - s - 6t + 15 = 0$

Solving these two equations simultaneously, we get:

$s = 3\quad t =  - 3$

Putting these into the general point formula above will give us the points of intersection. We have already found the direction vector for the perpendicular through the cross product. The following procedure to work out the equation need not be showed and thus is left to the reader.

Further Pure | Topic 6: Complex Numbers

Curriculum Heading

• understand De Moivre's theorem, for a positive integral exponent, in terms of geometrical effect of multiplication of complex number
• prove De Moivre's theorem for a positive integral exponent
• use de Moivre’s theorem for positive integral exponent to express trigonometrical ratios of multiple angles in terms of powers of trigonometrical ratios of the fundamental angle
• use de Moivre’s theorem, for a positive or negative rational exponent: in expressing powers of sin θ and cos θ in terms of multiple angles; in the summation of series; and in finding and using the nth roots of unity.

Introduction to Complex Number

Note:

$ \binom{n}{k} = \frac{n}{{k!(n - k)!}}$

$cis\,x = \cos x + \iota\sin x $ 

While it is assumed that one is completely well versed in the Complex Numbers Unit of the P3 CIE A Levels Mathematics paper, a brief summary of major points of note, is given below.

We define a complex number as two real numbers, a and b, given in the form: $a + \iota b$, where ${\iota ^2} =  - 1$. If b = 0, the complex number is wholly real, if a = 0, the number is wholly imaginary. The complex number is zero when $a = b = 0$.

Ex. Factorise:

1) ${x^2} + 1$

${x^2} + 1 = {x^2} - ( - 1) = {x^2} - {\iota ^2}$

$(x + \iota )(x - \iota )$

2) ${(x + a)^2} + {b^2}$

${(x + a)^2} - ( - {b^2}) = {(x + a)^2} + {\left( {\iota b} \right)^2}$

$ = (x + a + \iota b)(x + a - \iota b)$

We define the addition and multiplication as follows:

$(a + \iota b) + (c + \iota d) = (a + c) + \iota (b + d)$

$(a + \iota b)(c + \iota d) = (ac - bd) + \iota (ad + bc)$

The conjugate of a complex number, $z = a + \iota b$, is $\bar z = a - \iota b$. ${z^ * }$ is also a common notation for the conjugate.

We also know:

(i) $\overline {\left( {{z_1} + {z_2}} \right)}  = \overline {{z_1}}  + \overline {{z_2}} $

(ii) $\overline {{z_1}{z_2}} = \overline {{z_1}} \times \overline {{z_2}} $ 

(iii) $\overline {\left( {\frac{{{z_1}}}{{{z_2}}}} \right)}  = \frac{{\overline {{z_1}} }}{{\overline {{z_2}} }}$

(iv) $z\bar z = {\left| z \right|^2}$

(v) $z + \bar z = 2a$

(vi) $z - \bar z = 2b$

The Modulus and Argument of a Complex Number

For a complex number $z = a + \iota b$, the modulus and argument of the complex number is given by:

Modulus of z = r = $\left| z \right| = \sqrt {{a^2} + {b^2}} $

Argument of z = $\arg (z) = \theta  = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)$

Two alternate forms of the complex numbers are then:

$z = a + \iota b = r(\cos \theta  + \iota \sin \theta ) = r{e^{\iota \theta }}$

The following properties are relevant to our study:

Let ${z_1} = a + \iota b = {r_1}{e^{\iota {\theta _1}}}$ and ${z_2} = c + \iota d = {r_2}{e^{\iota {\theta _2}}}$

(i) $\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right| = {r_1}{r_2}$

(ii) $\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = \frac{{{r_1}}}{{{r_2}}}$

(iii) $\arg ({z_1}{z_2}) = {\theta _1} + {\theta _2}$

(iv) $\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = {\theta _1} - {\theta _2}$
(v) $\left| {{z_1}} \right| - \left| {{z_2}} \right| \le \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$

De Moivre's Theorem

${(\cos \theta  + \iota \sin \theta )^n} = \cos \,(n\theta ) + \iota sin\,(n\theta )$, where n is any rational number (which includes fractions).

The proof is detailed in the post involving Mathematical Induction, and you are directed to learn it as you may be asked to reproduce it in your Pure paper.

Ex. Evaluate ${\left( {\frac{{\sqrt 3  - \iota }}{{\sqrt 3  + \iota }}} \right)^6}$

Sol.

$\sqrt 3  - \iota  = r(\cos \theta  + \iota \sin \theta )$
$r = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{( - 1)}^2}}  = 2$
$\theta  =  - \frac{\pi }{6}$

${\left( {\sqrt 3  - \iota } \right)^6} = {2^6}{\left( {\cos \left( { - \frac{\pi }{6}} \right) + \iota \sin \left( { - \frac{\pi }{6}} \right)} \right)^6}$
$ = {2^6}(\cos ( - \pi ) + \iota \sin ( - \pi )) =  - {2^6}$

${\left( {\frac{1}{{\sqrt 3  + 1}}} \right)^6} = {\left( {\frac{{\sqrt 3  - \iota }}{{3 + 1}}} \right)^6} = \frac{1}{{{4^6}}}{\left( {\sqrt 3  - \iota } \right)^6} =  - \frac{{{2^6}}}{{{4^6}}}$

${\left( {\frac{{\sqrt 3  - \iota }}{{\sqrt 3  + \iota }}} \right)^6} = \left( { - {2^6}} \right)\frac{{\left( { - {2^6}} \right)}}{{{4^6}}} = 1$

Ex. Simplify $\frac{{{{\left( {\cos \alpha  - \iota sin\alpha } \right)}^{11}}}}{{{{(\cos \beta  + \iota \sin \beta )}^9}}}$

Sol.

${\left( {\cos \alpha  - \iota sin\alpha } \right)^{11}} = {\left( {\cos ( - \alpha ) + \iota sin( - \alpha )} \right)^{11}} = \cos ( - 11\alpha ) + \iota \sin ( - 11\alpha )$

${\left( {\cos \beta  + \iota sin\beta } \right)^{ - 9}} = \cos ( - 9\beta ) + \iota sin( - 9\beta )$

$\frac{{{{\left( {\cos \alpha  - \iota sin\alpha } \right)}^{11}}}}{{{{(\cos \beta  + \iota \sin \beta )}^9}}} = \left[ {\cos ( - 11\alpha ) + \iota \sin ( - 11\alpha )} \right]\left[ {\cos ( - 9\beta ) + \iota \sin ( - 9\beta )} \right]$

$= \cos ( - 11\alpha  - 9\beta ) + \iota \sin ( - 11\alpha  - 9\beta ) = \cos (11\alpha  + 9\beta ) - \iota \sin (11\alpha  + 9\beta )$

Using de Moivre’s theorem for positive integral exponent to express trigonometrical ratios of multiple angles in terms of powers of trigonometrical ratios of the fundamental angle

This is an application of the de Moivre's theorem that you are required to prove. In the general case, we have, if n is odd. 

$\cos (n\theta ) = {\cos ^n}\theta  - \binom{n}{2}{\cos ^{n - 2}}\theta {\sin ^2}\theta  + \binom{n}{4}{\cos ^{n - 4}}\theta {\sin ^4} +...+ {( - 1)^{\frac{{n - 1}}{2}}}\cos \theta {\sin ^{n - 1}}\theta $

$\sin (n\theta ) = \binom{n}{1}{\cos ^{n - 1}}\theta \sin \theta  - \binom{n}{3}{\cos ^{n - 3}}\theta {\sin ^3}\theta  + \binom{n}{5}{\cos ^{n - 5}}\theta {\sin ^5}\theta  - ... + {( - 1)^{\frac{{n - 1}}{2}}}{\sin ^n}\theta $

If n is even, we have:

$\cos (n\theta ) = {\cos ^n}\theta  - \binom{n}{2}{\cos ^{n - 2}}\theta {\sin ^2}\theta  + \binom{n}{4}{\cos ^{n - 4}}\theta {\sin ^4} + ... + {( - 1)^{\frac{n}{2}}}{\sin ^n}\theta $

$\sin (n\theta ) = \binom{n}{1}{\cos ^{n - 1}}\theta \sin \theta  - \binom{n}{3}{\cos ^{n - 3}}\theta {\sin ^3}\theta  + \binom{n}{5}{\cos ^{n - 5}}\theta {\sin ^5}\theta  - ... + {( - 1)^{\frac{{n - 2}}{2}}}n{\kern 1pt} cos\theta {\sin ^n}\theta $

As I said earlier, you will be required to prove these relationships, not memorise them. The examiner will expect the complete working I will exhibit in the next example. The general cases noted above are simply for you to tally your answers.


Ex. Express $\cos(6\theta)$ and $\frac{{\sin (6\theta )}}{{\sin \theta }}$ in terms of powers of $\cos(\theta)$


Sol.

Consider $z = \cos (6\theta ) + \iota \sin (6\theta )$

Then, by de Moivre's Theorem:

$z = {\left( {\cos (\theta ) + \iota \sin (\theta )} \right)^6}$

By the application of the binomial theorem we have:

$ = {\cos ^6}\theta  + 6{\cos ^5}\theta \iota \sin \theta  - 15co{s^4}\theta {\sin ^2}\theta  - 20{\cos ^3}\iota {\sin ^3}\theta  + 15{\cos ^2}\theta {\sin ^4}\theta  + 6\cos \theta \iota {\sin ^5}\theta  - {\sin ^6}\theta $

$\cos (6\theta ) + \iota \sin (6\theta ) = {\cos ^6}\theta  + 6{\cos ^5}\theta \iota \sin \theta  - 15co{s^4}\theta {\sin ^2}\theta  - 20{\cos ^3}\iota {\sin ^3}\theta  + 15{\cos ^2}\theta {\sin ^4}\theta  + 6\cos \theta \iota {\sin ^5}\theta  - {\sin ^6}\theta $

For two complex numbers to be equal, their real and imaginary parts must be equal as well. Therefore we can equate $\cos (6\theta)$ to all the terms with iota on the right hand side, i.e:

$\cos (6\theta ) = {\cos ^6}\theta  - 15{\cos ^4}\theta {\sin ^2}\theta  + 15{\cos ^2}\theta {\sin ^4}\theta  - {\sin ^6}\theta $

Now we substitute, ${\sin ^2}\theta  = 1 - {\cos ^2}\theta $

$\cos (6\theta ) = {\cos ^6}\theta  - 15{\cos ^4}\theta (1 - {\cos ^2}\theta ) + 15{\cos ^2}\theta {(1 - {\cos ^2}\theta )^2} - {(1 - {\cos ^2}\theta )^3}$

$\cos (6\theta ) = {\cos ^6}\theta  - 15{\cos ^4}\theta (1 - {\cos ^2}\theta ) + 15{\cos ^2}\theta {(1 - {\cos ^2}\theta )^2} - {(1 - {\cos ^2}\theta )^3}$

$ = {\cos ^6}\theta  - 15{\cos ^4}\theta  + 15{\cos ^6}\theta  + 15{\cos ^2}\theta (1 - 2{\cos ^2}\theta  + {\cos ^4}\theta ) - (1 - 3{\cos ^2}\theta  + 3{\cos ^4}\theta  - {\cos ^6}\theta )$

$ = {\cos ^6}\theta  - 15{\cos ^4}\theta  + 15{\cos ^6}\theta  + 15{\cos ^2}\theta  - 20{\cos ^4}\theta  + 15{\cos ^6}\theta  - 1 + 3{\cos ^2}\theta  - 3{\cos ^4}\theta  + {\cos ^6}\theta $

$ = 32{\cos ^6}\theta  - 48{\cos ^4}\theta  + 18{\cos ^2}\theta  - 1$

Following similar reasoning, one can aver the following to be true as well, by consider the imaginary parts on either side:

$\sin (6\theta ) = 6co{s^5}\theta \sin \theta  - 20{\cos ^3}\theta {\sin ^3}\theta  + 6\cos \theta {\sin ^5}\theta $

The iota is cancelled and then either side are divided by sin $\theta $.

$\frac{{\sin (6\theta )}}{{\sin \theta }} = 6{\cos ^5}\theta  - 20{\cos ^3}\theta {\sin ^2}\theta  + 6\cos \theta si{n^4}\theta $
$ \begin{array}{l} \frac{{\sin (6\theta )}}{{\sin \theta }} = 6{\cos ^5}\theta - 20{\cos ^3}\theta (1 - {\cos ^2}\theta ) + 6\cos \theta (1 - 2{\cos ^2}\theta + {\cos ^4}\theta )\\ = 6co{s^5}\theta - 20{\cos ^3}\theta + 20{\cos ^5}\theta + 6\cos \theta - 12co{s^3}\theta + 6{\cos ^5}\theta \\ = 32{\cos ^5}\theta - 32{\cos ^3}\theta + 6\cos \theta \end{array} $

Expressing powers of sin θ and cos θ in terms of multiple angles

Another application of the de Moivre's theorem takes us in an entirely opposite direction, i.e. we can express powers of the trigonometric ratios as sums containing terms of the form $sin(n\theta)$ or $cos(n\theta)$.

Consider:

${z^n} = {(\cos \theta  + \iota \sin \theta )^n} = \cos (n\theta ) + \iota \sin (n\theta )$
$\frac{1}{{{z^n}}} = {(\cos \theta  + \iota \sin \theta )^{ - n}} = \cos ( - n\theta ) + \iota \sin ( - n\theta ) = cos(n\theta ) - \iota \sin (n\theta )$

It is not difficult to see then:

$\begin{array}{l} {z^n} + \frac{1}{{{z^n}}} = 2\cos (n\theta )\\ {z^n} - \frac{1}{{{z^n}}} = 2\iota \sin (n\theta ) \end{array}$

Say I wished to express ${\cos ^3}\theta $ thusly. Outlined below, is the procedure to be followed.

$\begin{array}{l} 2\cos (n\theta ) = {z^n} + \frac{1}{{{z^n}}}\\ 2\cos \theta = z + \frac{1}{z}\\ 8{\cos ^3}\theta = {\left( {z + \frac{1}{z}} \right)^3}\\ = {z^3} + 3z + \frac{3}{z} + \frac{1}{{{z^3}}}\\ = \left( {{z^3} + \frac{1}{{{z^3}}}} \right) + 3\left( {z + \frac{1}{z}} \right)\\ = 2\cos (3\theta ) + 3(2\cos \theta ) \end{array}$

And thus, by using the binomial theorem and substituting in the general result we began with, we can conclude that:

${\cos ^3}\theta  = {\textstyle{1 \over 4}}\cos (3\theta ) + {\textstyle{3 \over 4}}\cos \theta $

Let us now consider sin theta raised to the power 5.

$\begin{array}{l} 2\iota \sin (n\theta ) = {z^n} - \frac{1}{{{z^n}}}\\ 2\iota \sin \theta = z - \frac{1}{z}\\ 32\iota {\sin ^5}\theta = {\left( {z - \frac{1}{z}} \right)^5}\\ = {z^5} + 5{\left( z \right)^4}\left( { - \frac{1}{z}} \right) + 10{(z)^3}{\left( { - \frac{1}{z}} \right)^2} + 10{(z)^2}{\left( { - \frac{1}{z}} \right)^3} + 5z{\left( { - \frac{1}{z}} \right)^4} + {\left( { - \frac{1}{z}} \right)^5}\\ = {z^5} - \frac{1}{{{z^5}}} - 5\left( {{z^3} - \frac{1}{{{z^3}}}} \right) + 10\left( {z - \frac{1}{z}} \right)\\ = 2\iota \sin (5\theta ) - 10\iota \sin (3\theta ) + 20\iota sin\theta \end{array}$

Thus, we can say with certainty, that:

${\sin ^5}\theta  = {\textstyle{1 \over {16}}}\sin (5\theta ) - {\textstyle{5 \over {16}}}\sin (3\theta ) + {\textstyle{5 \over 8}}\sin \theta $

Summation of series using de Moivre's Theorem

There is no general method to be followed here that may be taught. What follows are examples from various sources, of various sorts, to familiarise the reader with the points worthy of noting in such questions.

The first of these example is little more difficult than the usual questions on the topic. You would probably not get this on an examination, but having mastered this, the topic becomes significantly easier. The only surety here is that all the information you will require will be present either in this blog, or the question itself.

Ex. Prove $1 + {e^{\iota \theta }} = 2\cos \left( {\frac{\theta }{2}} \right){e^{\frac{{\iota \theta }}{2}}}$

Consider:

$\begin{array}{l} 1 = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{ - \iota \theta }}{2}}}\\ {e^{\iota \theta }} = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{\iota \theta }}{2}}}\\ 1 + {e^{\iota \theta }} = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{ - \iota \theta }}{2}}} + {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{\iota \theta }}{2}}}\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {{e^{\frac{{\iota \theta }}{2}}} + {e^{ - \frac{{\iota \theta }}{2}}}} \right)\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {\cos \left( {\frac{\theta }{2}} \right) + \iota \sin \left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right) - \iota \sin \left( {\frac{\theta }{2}} \right)} \right)\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {2\cos \left( {\frac{\theta }{2}} \right)} \right) \end{array}$

Note: I have used the following identity $\sin ( - \theta ) =  - sin(\theta )$

Find now the sum the series:

$C = 1 + \binom{n}{1}\cos \theta  + \binom{n}{2}\cos (2\theta ) + ... + \binom{n}{n}\cos (n\theta )$

To sum $C = \sum\limits_{k = 0}^n {\binom{n}{k}\cos (k\theta )} $  (This is equivalent to what was written above. If you believe, your command of the sigma notation is not up to par, you are directed to read up the chapter on summation of series before this section), we consider the following sum:

$ S = \sum\limits_{k = 0}^n {\binom{n}{k}sin(k\theta )} $

Now consider the following summation:

$C + \iota S = 1 + \binom{n}{1}\cos (\theta ) + \binom{n}{2}cos(2\theta ) + ... + \binom{n}{n}cos(n\theta ) + \binom{n}{1}\iota sin(\theta ) + \binom{n}{2}\iota sin(2\theta ) + ... + \binom{n}{n}\iota \sin (n\theta )$
$ = 1 + \binom{n}{1}\left( {cos\theta  + \iota \sin \theta } \right) + \binom{n}{2}\left( {\cos (2\theta ) + \iota \sin (2\theta )} \right) + ... + \binom{n}{n}\left( {\cos (n\theta ) + \iota \sin (n\theta )} \right)$
$ = 1 + \binom{n}{1}{e^{\iota \theta }} + \binom{n}{2}{e^{2\iota \theta }} + ... + \binom{n}{n}{e^{n\iota \theta }}$
$ = 1 + \binom{n}{1}{e^{\iota \theta }} + \binom{n}{2}{\left( {{e^{\iota \theta }}} \right)^2} + ... + \binom{n}{n}{\left( {{e^{\iota \theta }}} \right)^n}$

We recognise this as a binomial expansion

$ = {\left( {1 + {e^{\iota \theta }}} \right)^n}$

It is now that we use the initial result:

$\begin{array}{l} C + \iota S = {\left( {2\cos \left( {\frac{\theta }{2}} \right)\left( {{e^{\frac{{\iota \theta }}{2}}}} \right)} \right)^n}\\ = {2^n}{\cos ^n}\left( {\frac{\theta }{2}} \right)\left( {{e^{\frac{{n\iota \theta }}{2}}}} \right)\\ = {2^n}{\cos ^n}\frac{\theta }{2}\left( {\cos \left( {\frac{{n\theta }}{2}} \right) + \iota \sin \left( {\frac{{n\theta }}{2}} \right)} \right) \end{array}$

Now we recall that for two complex numbers to be equal, the real parts and the imaginary parts should be equal individually, on either sides.

We can therefore conclude:

$C = \sum\limits_{k = 0}^n {\binom{n}{k}\cos (k\theta )}  = {2^n}{\cos ^n}\left( {\frac{\theta }{2}} \right)\cos \left( {\frac{{n\theta }}{2}} \right)$

$S = \sum\limits_{k = 0}^n {\binom{n}{k}sin(k\theta )}  = {2^n}{\cos ^n}\left( {\frac{\theta }{2}} \right)\sin \left( {\frac{{n\theta }}{2}} \right)$

Ex. Sum the following series:

$S = \sin (\alpha ) + k\sin (\alpha  + \beta ) + k\sin (\alpha  + 2\beta ) + ...$, where $\left| k \right| < 1$

We consider the following sum:

$C = \cos (\alpha ) + k\cos (\alpha  + \beta ) + k\cos (\alpha  + 2\beta ) + ...$

$\begin{array}{l} C + \iota S = (\cos (\alpha ) + k\cos (\alpha + \beta ) + {k^2}\cos (a + 2\beta ) + ...) + \iota (\sin (\alpha ) + k\sin (\alpha + \beta ) + {k^2}\sin (\alpha + 2\beta ) + ...)\\ = (\cos (\alpha ) + \iota \sin (a)) + k(\cos (\alpha + \beta ) + \iota sin(\alpha + \beta )) + {k^2}(\cos (a + 2\beta ) + \iota \sin (\alpha + 2\beta ) + ...)\\ = {e^{\iota \alpha }} + k{e^{\iota (\alpha + \beta )}} + {k^2}{e^{\iota (\alpha + 2\beta )}} + ...\\ = {e^{\iota \alpha }} + k{e^{\iota \alpha }}{e^{\iota \beta }} + {k^2}{e^{\iota \alpha }}{\left( {{e^{\iota \beta }}} \right)^2} + ... \end{array}$

For k is less than one, and this is a geometric series, the sum to infinity formula is relevant here. Also note that the required sum is the imaginary part of this sum, therefore:

$Initial\,Term = {e^{\iota \alpha }}\quad r = k{e^{\iota \beta }}$
$\begin{array}{l} {S_\infty } = {\mathop{\rm Im}\nolimits} \left[ {\frac{a}{{1 - r}}} \right] = {\mathop{\rm Im}\nolimits} \left[ {\frac{{{e^{\iota \alpha }}}}{{1 - k.{e^{\iota \beta }}}}} \right]\\ = {\mathop{\rm Im}\nolimits} \left[ {\frac{{\cos (\alpha ) + \iota \sin (\alpha )}}{{1 - k\cos (\beta ) - k\iota \sin (\beta )}} \times \frac{{1 - k\cos (\beta ) + \iota \sin (\alpha )}}{{1 - k\cos (\beta ) + k\iota sin(\beta )}}} \right]\\ = {\mathop{\rm Im}\nolimits} \left[ {\frac{{\cos (\alpha ) + \iota \sin (\alpha )\left[ {1 - k\cos (\beta ) + \iota \sin (\alpha )} \right]}}{{{{(1 - k\cos (\beta ))}^2} + {{(k\sin (\beta ))}^2}}}} \right]\\ = {\mathop{\rm Im}\nolimits} \left[ {\frac{{\cos (\alpha ) + \iota \sin (\alpha )\left[ {1 - k\cos (\beta ) + \iota \sin (\alpha )} \right]}}{{1 - 2k\cos (\beta ) + {k^2}{{\cos }^2}\beta + {k^2}{{\sin }^2}\beta }}} \right] \end{array}$

The subsequent algebra is not difficult, wherewith we arrive at the answer:

${S_\infty } = \frac{{\sin \alpha (1 - k\cos \beta ) + k\cos \alpha sin\beta }}{{1 - 2k\cos \beta  + {k^2}}}$

Ex. Sum the infinite series:

$C = 1 + \frac{1}{2}\cos (2\theta ) - \frac{1}{{2 \cdot 4}}\cos (4\theta ) + \frac{{1 \cdot 3}}{{2 \cdot 4 \cdot 6}}\cos (6\theta ) + ...$

$S = \frac{1}{2}sin(2\theta ) - \frac{1}{{2 \cdot 4}}sin(4\theta ) + \frac{{1 \cdot 3}}{{2 \cdot 4 \cdot 6}}sin(6\theta ) + ...$

Again:

$\begin{array}{l} C + \iota S = 1 + \frac{1}{2}(\cos (2\theta ) + \iota \sin (2\theta )) - \frac{1}{{2 \cdot 4}}(\cos (4\theta ) + \iota \sin (4\theta )) + \frac{{1 \cdot 3}}{{2 \cdot 4 \cdot 6}}(\cos (6\theta ) + \iota \sin (6\theta )) + ...\\ = 1 + \frac{1}{2}{e^{2\iota \theta }} + \frac{1}{8}{\left( {{e^{2\iota \theta }}} \right)^2} + \frac{1}{{16}}{\left( {{e^{2\iota \theta }}} \right)^3}...\\ = {(1 + {e^{2\iota \theta }})^{\frac{1}{2}}}\\ = {\left( {1 + \cos (2\theta ) + \iota sin(2\theta )} \right)^{\frac{1}{2}}}\\ = {(2co{s^2}(\theta ) + 2\iota sin(\theta )cos(\theta ))^{\frac{1}{2}}}\\ = \sqrt {2\cos \theta } {\left( {\cos (\theta ) + \iota \sin (\theta )} \right)^{\frac{1}{2}}} \end{array}$

By applying the de Moivre's theorem, we can give the finally answer:

$C = \sqrt {2\cos \theta } \left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$

Ex. (May/June 2008, Paper 01, Question 10)

By considering $\sum\limits_{n = 1}^N {{z^{2n - 1}}} $, where $z = {e^{\iota \theta }}$, show that:

$\sum\limits_{n = 1}^N {\cos (2n - 1)\theta }  = \frac{{\sin (2N\theta )}}{{2\sin \theta }}$.

We consider what is given to us in the question:

$\begin{array}{l} \sum\limits_{n = 1}^N {{z^{2n - 1}}} = z + {z^3} + {z^5}...\\ = z + z({z^2}) + {z^3}({z^2})... \end{array}$

$z = \cos \theta  + \iota \sin \theta $

It is a geometric progression, the sum is then given by:

$\begin{array}{l} S = \frac{{a(1 - {r^n})}}{{1 - r}} = \frac{{z\left( {1 - {{\left( {{z^2}} \right)}^n}} \right)}}{{1 - {z^2}}}\\ = \frac{{z - {z^{2N + 1}}}}{{1 - {z^2}}} \end{array}$

In the denominator we substitute the value of z, and rationalise by multiplying with the conjugate.

$\begin{array}{l} = \frac{{z - {z^{2N + 1}}}}{{1 - \cos (2\theta ) - \iota sin(2\theta )}} \times \frac{{1 - \cos (2\theta ) + \iota \sin (2\theta )}}{{1 - \cos (2\theta ) + \iota \sin (2\theta )}}\\ = \frac{{z - {z^{2N + 1}}}}{{{{(1 - \cos (2\theta ))}^2} + ({{\sin }^2}(2\theta ))}} \times 1 - \cos (2\theta ) + \iota \sin (2\theta )\\ = \frac{{z - {z^{2N + 1}}}}{{1 - 2\cos (2\theta ) + co{s^2}(2\theta ) + {{\sin }^2}(2\theta )}} \times 1 - \cos (2\theta ) + \iota \sin (2\theta )\\ = \frac{{z - {z^{2N + 1}}}}{{2 - 2\cos (2\theta )}} \times 1 - \cos (2\theta ) + \iota \sin (2\theta ) \end{array}$

Now you may ask my reasoning behind not substituting in the numerator. My answer would be that it makes the algebra somewhat easier.

But before consider the following:

$\begin{array}{*{20}{l}} {1 - \overline {{z^2}} ,\;\bar z\;is\;the\;conjugate\;of\;z}\\ { = 1 - \cos (2\theta ) + sin(2\theta )}\\ { = 1 - \left[ {cos(2\theta ) - sin(2\theta )} \right]}\\ {As:}\\ {\cos (x) = \cos ( - x)}\\ { - \sin (x) = \sin ( - x)}\\ {1 - \left[ {\cos ( - 2\theta ) + \iota \sin ( - 2\theta )} \right]}\\ {1 - \left[ {{{\left( {\cos (\theta ) + \iota sin(\theta )} \right)}^{ - 2}}} \right]}\\ {1 - {z^{ - 2}}} \end{array}$

Using this I can write:

$\begin{array}{l} S = \frac{{z - {z^{2N + 1}}}}{{2 - 2\cos (2\theta )}} \times 1 - \cos (2\theta ) + \iota \sin (2\theta )\\ = \frac{{z - {z^{2N + 1}}}}{{2 - 2\cos (2\theta )}} \times 1 - {z^{ - 2}}\\ = \frac{{(z - {z^{2N + 1}})(1 - {z^{ - 2}})}}{{2 - 2\cos (2\theta )}}\\ = \frac{{z - {z^{ - 1}} - {z^{2N + 1}} + {z^{2N - 1}}}}{{2 - 2\cos (2\theta )}} \end{array}$

Now remember that we only require the real part of this summation:

$\begin{array}{l} {\mathop{\rm Re}\nolimits} \left[ {\frac{{z - {z^{ - 1}} - {z^{2N + 1}} + {z^{2N - 1}}}}{{2 - 2\cos (2\theta )}}} \right] = \frac{{\cos (2N - 1)\theta - cos(2N + 1)\theta }}{{2 - 2\cos (2\theta )}}\\ = \frac{{\cos (2N\theta )\cos (\theta ) + sin(2N\theta )sin(\theta ) - cos(2N\theta )cos(\theta ) + sin(2N\theta )sin(\theta )}}{{2 - 2(1 - 2{{\sin }^2}\theta )}}\\ = \frac{{2\sin (2N\theta )sin\theta }}{{4{{\sin }^2}\theta }}\\ = \frac{{\sin (2N\theta )}}{{2\sin \theta }} \end{array}$

Deduce that:

$\sum\limits_{n = 1}^N {\left( {2n - 1} \right)\sin \left[ {\frac{{\left( {2n - 1} \right)\pi }}{N}} \right]}  =  - N{\mathop{\rm cosec}\nolimits} \frac{\pi }{N}$

This is not a particularly difficult question, though every student I had the pleasure to know, has unfailingly stumped by this question. The only thing to be kept in mind is what we have already proved, and what is left to be proved, i.e.

$\cos \theta  + \cos (3\theta ) + \cos (5\theta ) + ... + cos\left[ {\left( {2N - 1} \right)\theta } \right] = \frac{{\sin (2N\theta )}}{{2\sin \theta }}$

$\sin \theta  + 3\sin (3\theta ) + 5\sin (5\theta ) + ... + \left( {2N - 1} \right)\sin \left[ {\left( {2N - 1} \right)\theta } \right] =  - N{\mathop{\rm cosec}\nolimits} \frac{\pi }{N}$

By now it must be apparent that the required sum is the derivative of the proven sum, with respect to $\theta$. Differentiate either side and then substitute:  $\theta  = \frac{\pi }{N}$

Nth roots of complex numbers

Let us begin this section, by considering a very specific example of the overall topic, say, the cube root of unity. Which is simply:

$x = \sqrt[3]{1}$

To which you will most likely reply that the answer is 1, while effectively dismissing the fact that this is a cubic equation, and should have, by the fundamental theorem of algebra, three roots ( accounting for multiplicity or repeated roots). We find the other roots by the remainder theorem.

By simple calculation we know that x - 1, is a root of this equation. Further we recall, that provided n is odd:

$\begin{array}{l} {x^n} - {a^n} = \left( {x - a} \right)\left( {{x^{n - 1}} + a{x^{n - 2}} + {a^2}{x^{n - 3}} + ... + {a^{n - 1}}} \right)\\ {x^n} + {a^n} = \left( {x - a} \right)\left( {{x^{n - 1}} - a{x^{n - 2}} + {a^2}{x^{n - 3}} - ... + {a^{n - 1}}} \right) \end{array}$

By this we can write:

${x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)$

Factorising $x^2 + x + 1$, we get:

$x = \frac{{ - 1 \pm \iota \sqrt 3 }}{2}$

If we let one of these roots be w, then its square will be equal to the other, which is to say

${\left( {\frac{{ - 1 + \iota \sqrt 3 }}{2}} \right)^2} = \frac{{ - 1 - \iota \sqrt 3 }}{2}$

The roots are therefore 1, w, w^2.

There is an another method to achieve the same result, with slightly greater ease, and it is the subject of consideration now. The superiority of this method is more pronounced in equations of degrees, greater than 3. 

By taking 1 to be the complex number $1 + \iota (0)$, and converting it to the cis x form by finding the modulus and the argument of the complex number we get:

$1 = \cos (0) + \iota \sin (0)$

It follows that;

$\begin{array}{l} {x^3} = \cos (0) + \iota \sin (0)\\ x = {\left( {\cos (0) + \iota \sin (0)} \right)^{\frac{1}{3}}} = \cos \left( {\frac{0}{3}} \right) + \iota \sin \left( {\frac{0}{3}} \right)\\ x = \cos (0) + \iota \sin (0) = 1 \end{array}$

Consider now the periodicity of the trigonometric function, which is to say:

$\cos (0) + \iota sin(0) = \cos (0 + 2\pi ) + \iota sin(0 + 2\pi ) = \cos \left( {0 + 2\left( {2\pi } \right)} \right) + \iota sin\left( {0 + 2\left( {2\pi } \right)} \right)$

By this fact, I can say:

$\begin{array}{l} {x^3} = \cos (2\pi ) + \iota sin(2\pi )\\ x = {\left( {\cos (2\pi ) + \iota sin(2\pi )} \right)^{\frac{1}{3}}}\\ = \cos \left( {\frac{{2\pi }}{3}} \right) + \iota \sin \left( {\frac{{2\pi }}{3}} \right)\\ = - \frac{1}{2} + \iota\frac{{\sqrt 3 }}{2} \end{array}$

Also we have:

$\begin{array}{l} {x^3} = \cos (4\pi ) + \iota sin(4\pi )\\ x = {\left( {\cos (4\pi ) + \iota sin(4\pi )} \right)^{\frac{1}{3}}}\\ = \cos \left( {\frac{{4\pi }}{3}} \right) + \iota \sin \left( {\frac{{4\pi }}{3}} \right)\\ = - \frac{1}{2} - \iota\frac{{\sqrt 3 }}{2} \end{array}$

Obviously the method is sound. Also note:

${\left( {\cos \left( {\frac{{2\pi }}{3}} \right) + \iota \sin \left( {\frac{{2\pi }}{3}} \right)} \right)^2} = \cos \left( {\frac{{4\pi }}{3}} \right) + \iota \sin \left( {\frac{{4\pi }}{3}} \right) = {\left( {\frac{{ - 1 + \iota \sqrt 3 }}{2}} \right)^2} = \frac{{ - 1 - \iota \sqrt 3 }}{2}$

If you will use $6\pi$, the answer you get will be 1,  if you used $8\pi$ you will get w. You will continue to cycle through these three roots, as you take greater multiples of $2\pi$. This method obviously is not restricted to unity.

Let me then generalise the method, and furnish the reader with more examples.

For the equation:

${z^n} = a$

If $a \ne 0$, and n is a positive integer, then there are exactly n distinct complex numbers ${z_0},{z_1},...,{z_{n - 1}}$. Furthermore, these roots can be obtained by the formula:

${z_k} = {\left| a \right|^{\frac{1}{n}}}\left[ {\cos \left( {{\phi _k}} \right) + \iota \sin \left( {{\phi _k}} \right)} \right]$

where 

${\phi _k} = \frac{{\arg \left( a \right) + 2k\pi }}{n}$

Note that all the roots have the same modulus, and have an argument differing by multiples of $2\pi$. The roots of a complex number actually form a geometric progression:

For:

$\begin{array}{l} {z_k} = {\left| a \right|^{\frac{1}{n}}}{\mathop{\rm cis}\nolimits} \left( {\frac{{\arg \left( a \right) + 2k\pi }}{n}} \right)\\ {z_{k + 1}} = {\left| a \right|^{\frac{1}{n}}}{\mathop{\rm cis}\nolimits} \left( {\frac{{\arg \left( a \right) + 2\left( {k + 1} \right)\pi }}{n}} \right)\\ = {\left| a \right|^{\frac{1}{n}}}{\mathop{\rm cis}\nolimits} \left( {\frac{{\arg \left( a \right) + 2k\pi }}{n} + \frac{{2\pi }}{n}} \right)\\ = {\left| a \right|^{\frac{1}{n}}}{\mathop{\rm cis}\nolimits} \left( {\frac{{\arg \left( a \right) + 2k\pi }}{n}} \right)cis\left( {\frac{{2\pi }}{n}} \right) = {z_k}cis\left( {\frac{{2\pi }}{n}} \right) \end{array}$

Ex. Find the cube roots of 64.

Sol.

You can use either of the two method detailed above but the point to be noticed here, is that you can deduce the result, by the use of the cube roots of unity. The method I will outline here stands for any integral value of n, where the number whose roots to be found are real. I recall a question CIE posed on the matter, though of what paper, I have forgotten. The general method is such:

$\begin{array}{l} {z^3} = 64\\ z = \sqrt[3]{{64}} = \sqrt[3]{{64 \times 1}}\\ z = \sqrt[3]{{64}}\sqrt[3]{1} \end{array}$

Now convert unity into the complex form:

$\begin{array}{l} {z^3} = 64\\ z = \sqrt[3]{{64}} = \sqrt[3]{{64 \times 1}}\\ z = \sqrt[3]{{64}}\left[ {\cos \left( {\frac{{2k\pi }}{3}} \right) + \iota \sin \left( {\frac{{2k\pi }}{3}} \right)} \right] = 4\left[ {\cos \left( {\frac{{2k\pi }}{3}} \right) + \iota \sin \left( {\frac{{2k\pi }}{3}} \right)} \right] \end{array}$

This suggests that the cube roots of 64, may be found by multiplying the roots of unity by the 4 ( which is sometimes referred to as the principal root). 

Hence the roots are:

$\sqrt[3]{{64}} = 4;\;4\left( {\frac{{ - 1 + \sqrt 3 }}{2}} \right);\;4\left( {\frac{{ - 1 - \sqrt 3 }}{2}} \right)$

Now let us solve some examination questions:

Ex. (October/November 2013, Paper 12, Question 11 Either)

State the fifth roots of unity in the form $\cos \theta  + \iota \sin \theta $, where $ - \pi \le \theta \le \pi $

Obviously 1 is one roots; the rest can thusly be noted:

$\sqrt[5]{1} = \cos \left( {\frac{{2k\pi }}{5}} \right) + \iota \sin \left( {\frac{{2k\pi }}{5}} \right)$

Choice of k here is worth elaborating upon. The question has imposed upon us a range, wherefore the value of theta must be less or equal than pi. For this reason the value of 3 and 4 for the value of k are ignored. Instead the range also us to use negative values of k. Therefore the values of k that we shall use to obtain the roots are: -2, -1, 0, 1, 2.

Simplify:

$\left( {x - \left[ {\cos {\textstyle{2 \over 5}}\pi  + \iota \sin {\textstyle{2 \over 5}}\pi } \right]} \right)\left( {x - \left[ {\cos {\textstyle{2 \over 5}}\pi  - \iota \sin {\textstyle{2 \over 5}}\pi } \right]} \right)$ 

Sol.

$\begin{array}{l} \left( {x - \left[ {\cos {\textstyle{2 \over 5}}\pi + \iota \sin {\textstyle{2 \over 5}}\pi } \right]} \right)\left( {x - \left[ {\cos {\textstyle{2 \over 5}}\pi - \iota \sin {\textstyle{2 \over 5}}\pi } \right]} \right)\\ = {x^2} - \left[ {\cos {\textstyle{2 \over 5}}\pi - \iota \sin {\textstyle{2 \over 5}}\pi } \right]x - \left[ {\cos {\textstyle{2 \over 5}}\pi + \iota \sin {\textstyle{2 \over 5}}\pi } \right]x + \left[ {{{\cos }^2}{\textstyle{2 \over 5}}\pi + {{\sin }^2}{\textstyle{2 \over 5}}\pi } \right]\\ = {x^2} - \left[ {2\cos {\textstyle{2 \over 5}}\pi } \right]x + 1 \end{array}$

Hence find the real factors of x^5 - 1:

We write x^5 - 1 as a product of its roots and then multiply the conjugate roots, in the manner shown above to arrive at the answer:

$\left( {{x^2} - \left[ {2\cos {\textstyle{2 \over 5}}\pi } \right]x + 1} \right)\left( {{x^2} - \left[ {2\cos {\textstyle{4 \over 5}}\pi } \right]x + 1} \right)\left( {x - 1} \right)$

Express the six roots of the equation  ${x^6} - {x^3} + 1 = 0$, as three conjugate pairs, in the form $\cos \theta  \pm \iota \sin \theta $

Here we revisit the method of substitution to solve such equations. We let:

${x^3} = y$

Hence:

${y^2} - y + 1 = 0$

Solving the quadratic, we get:

$\begin{array}{l} y = \frac{1}{2} \pm \iota \frac{{\sqrt 3 }}{2}\\ {x^3} = \frac{1}{2} + \iota \frac{{\sqrt 3 }}{2}\\ {x^3} = \cos \left( {\frac{\pi }{3} + 2k\pi } \right) \pm \iota \sin \left( {\frac{\pi }{3} + 2k\pi } \right)\\ x = \cos \left( {\frac{\pi }{9} + \frac{{2k\pi }}{3}} \right) \pm \iota \sin \left( {\frac{\pi }{9} + \frac{{2k\pi }}{3}} \right)\\ x = \cos \left( {\frac{\pi }{9}} \right) \pm \iota \sin \left( {\frac{\pi }{9}} \right);\;\cos \left( {\frac{{7\pi }}{9}} \right) \pm \iota \sin \left( {\frac{{7\pi }}{9} + } \right);\;\cos \left( {\frac{{13\pi }}{9}} \right) \pm \iota \sin \left( {\frac{{13\pi }}{9}} \right) \end{array}$

Using the roots of the quadratic, we found the roots of the original of equation given.

Hence find the real factors of the equation

Sol.

As with the previous one, we multiply the conjugates to get the real factors, the answer should be of the form:

${x^6} - {x^3} + 1 = \left( {x - 2\cos \frac{\pi }{9}x + 1} \right)\left( {x - 2\cos \frac{{7\pi }}{9}x + 1} \right)\left( {x - 2\cos \frac{{13\pi }}{9}x + 1} \right)$