Question

Use De Moivre’s theorem, to find the indicated power of ${\left( { - \sqrt 2 - \sqrt 2 i} \right)^5}$?

Answer 1

Hint: This problem is from complex numbers and we need to know De Moivre’s theorem to solve this question. First we have to consider the given equation and we have to equate it to the general form and then we have to consider the power and by De Moivre’s theorem, we need to solve this problem.

Complete step by step answer:
For any complex number, $\left( {\cos n\theta + i\sin n\theta } \right)$ is the value or one of the values for ${\left( {\cos \theta + i\sin \theta } \right)^n}$. Write the given equation in polar form i.e., as $r\left( {\cos \theta + i\sin \theta } \right)$ and equate the real and imaginary parts. First let’s consider the question without considering the power value which is,
$ \Rightarrow \left( { - \sqrt 2 - \sqrt 2 i} \right) = r\left( {\cos \theta + i\sin \theta } \right)$
$ \Rightarrow \left( { - \sqrt 2 - \sqrt 2 i} \right) = r\cos \theta + ri\sin \theta $
Equating real and imaginary term, we get,
$ \Rightarrow r\cos \theta = - \sqrt 2 .............\left( i \right)$
$ \Rightarrow r\sin \theta = - \sqrt 2 ...............\left( {ii} \right)$
Squaring and adding the $\left( i \right)$ and $\left( {ii} \right)$ equation we get,
$ \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = {\left( { - \sqrt 2 } \right)^2} + {\left( { - \sqrt 2 } \right)^2}$
$ \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 2 + 2$

On taking ${r^2}$ as a common term, we get
$ \Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = 2 + 2$
As we know ${\sin ^2}\theta + {\cos ^2}\theta = 1$, therefore, we get
$ \Rightarrow {r^2} = 4$
$ \Rightarrow r = 2$
Substituting the value of $r$ in equation $\left( i \right)$ and $\left( {ii} \right)$ we get,
From $\left( i \right)$, we have
$ \Rightarrow \cos \theta = \dfrac{{ - \sqrt 2 }}{2}$
$ \Rightarrow \cos \theta = \dfrac{{ - 1}}{{\sqrt 2 }}$
From $\left( {ii} \right)$, we have
$ \Rightarrow \sin \theta = \dfrac{{ - \sqrt 2 }}{2}$
$ \Rightarrow \sin \theta = \dfrac{{ - 1}}{{\sqrt 2 }}$

And the value of $\theta = \dfrac{{5\pi }}{4}$, for both, now let’s substitute the value of $r$ and $\theta $ in the equation
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = {\left( {2\cos \dfrac{{5\pi }}{4} + 2i\sin \dfrac{{5\pi }}{4}} \right)^5}$
On taking ${2^5}$ as a common term, we get
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = {2^5}{\left( {\cos \dfrac{{5\pi }}{4} + i\sin \dfrac{{5\pi }}{4}} \right)^5}$
According to DeMoivre’s theorem, ${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)$. Therefore, we get
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = {2^5}\left( {\cos \dfrac{{5 \times 5\pi }}{4} + i\sin \dfrac{{5 \times 5\pi }}{4}} \right)$
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = {2^5}\left( {\cos \dfrac{{25\pi }}{4} + i\sin \dfrac{{25\pi }}{4}} \right)$

This can also be written as,
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = {2^5}\left( {\cos \left( {6\pi + \dfrac{\pi }{4}} \right) + i\sin \left( {6\pi + \dfrac{\pi }{4}} \right)} \right)$
As we know $\cos \left( {2n\pi + \theta } \right) = \cos \theta $ and $\sin \left( {2n\pi + \theta } \right) = \sin \theta $, for all values of $\theta $ and $n \in N$. Therefore, we have
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = 32\left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right)$
As we know $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$. Therefore, we get
$ \Rightarrow {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = 32\left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right)$
On simplification, we get
$ \therefore {\left( { - \sqrt 2 - \sqrt 2 i} \right)^5} = 16\sqrt 2 + 16\sqrt 2 i$

Therefore, the indicated power of ${\left( { - \sqrt 2 - \sqrt 2 i} \right)^5}$ is $16\sqrt 2 + 16\sqrt 2 i$.

Note: A complex number is a combination of real and the imaginary number where $i$ represents the imaginary. To solve or simplify the complex number De Moivre’s theorem is used. It is stated as for any complex number $\theta $ we have, ${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)$ where $n$ is a positive integer and $i$ is the imaginary part, $i = \sqrt { - 1} $ and also ${i^2} = - 1$.Remember that this formula is not valid for non-integer powers $m$. The trigonometry ratios values for the standard angles are used to simplify further.