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.
- 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) $z - \bar z = 2b$
Note:
$ \binom{n}{k} = \frac{n}{{k!(n - k)!}}$
$cis\,x = \cos x + \iota\sin x $ 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) $\left| {{z_1}} \right| - \left| {{z_2}} \right| \le \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$
(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.
${\left( {\cos \alpha - \iota sin\alpha } \right)^{11}} = {\left( {\cos ( - \alpha ) + \iota sin( - \alpha )} \right)^{11}} = \cos ( - 11\alpha ) + \iota \sin ( - 11\alpha )$
${\left( {\cos \beta + \iota sin\beta } \right)^{ - 9}} = \cos ( - 9\beta ) + \iota sin( - 9\beta )$
$\frac{{{{\left( {\cos \alpha - \iota sin\alpha } \right)}^{11}}}}{{{{(\cos \beta + \iota \sin \beta )}^9}}} = \left[ {\cos ( - 11\alpha ) + \iota \sin ( - 11\alpha )} \right]\left[ {\cos ( - 9\beta ) + \iota \sin ( - 9\beta )} \right]$
$= \cos ( - 11\alpha - 9\beta ) + \iota \sin ( - 11\alpha - 9\beta ) = \cos (11\alpha + 9\beta ) - \iota \sin (11\alpha + 9\beta )$
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} $
$\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 $
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.
And thus, by using the binomial theorem and substituting in the general result we began with, we can conclude that:
Let us now consider sin theta raised to the power 5.
Thus, we can say with certainty, that:
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:
${x^6} - {x^3} + 1 = \left( {x - 2\cos \frac{\pi }{9}x + 1} \right)\left( {x - 2\cos \frac{{7\pi }}{9}x + 1} \right)\left( {x - 2\cos \frac{{13\pi }}{9}x + 1} \right)$
$\begin{array}{l} 1 = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{ - \iota \theta }}{2}}}\\ {e^{\iota \theta }} = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{\iota \theta }}{2}}}\\ 1 + {e^{\iota \theta }} = {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{ - \iota \theta }}{2}}} + {e^{\frac{{\iota \theta }}{2}}} \times {e^{\frac{{\iota \theta }}{2}}}\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {{e^{\frac{{\iota \theta }}{2}}} + {e^{ - \frac{{\iota \theta }}{2}}}} \right)\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {\cos \left( {\frac{\theta }{2}} \right) + \iota \sin \left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right) - \iota \sin \left( {\frac{\theta }{2}} \right)} \right)\\ = {e^{\frac{{\iota \theta }}{2}}}\left( {2\cos \left( {\frac{\theta }{2}} \right)} \right) \end{array}$
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:
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:
${x^6} - {x^3} + 1 = \left( {x - 2\cos \frac{\pi }{9}x + 1} \right)\left( {x - 2\cos \frac{{7\pi }}{9}x + 1} \right)\left( {x - 2\cos \frac{{13\pi }}{9}x + 1} \right)$
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