Chapter I, Problem 36E

To determine

To find: The 8^th power of the complex number $\left(1-i\right)$ by using De Moivre’s Theorem.

Answer to Problem 36E

The value of the complex number ${\left(1-i\right)}^8$ is $16$.

Explanation of Solution

Theorem used:

De Moivre’s Theorem:

If $z=r\left(\cos \theta +i\sin \theta \right)$ and $n$ be a positive integer then,

$z^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n=r^n\left(\cos n\theta +i\sin n\theta \right)$.

Calculation:

Rewrite the complex number $\left(1-i\right)$ in polar form.

The polar form of the complex number $z=a+bi$ is $z=r\left(\cos \theta +i\sin \theta \right)$ where $r=\left|z\right|=\sqrt{a^2+b^2}$  and $\tan \theta =\frac{b}{a}$.

Consider the complex number $\left(1-i\right)$.

Obtain the argument of the complex number $\left(1-i\right)$.

$\begin{array}{c}\tan \theta =\frac{-1}{1}\\ =-1\end{array}$

Thus, the argument of argument of the complex number $\left(1-i\right)$ is $\theta ={\tan }^{-1}\left(-1\right)=\frac{7\pi }{4}$

Obtain the modulus of the complex number $\left(1-i\right)$.

$\begin{array}{c}r=\left|\left(1-i\right)\right|=\\ =\sqrt{1^2+{\left(-1\right)}^2}\\ =\sqrt{1+1}\\ =\sqrt{2}\end{array}$

Thus, the value of $r=\sqrt{2}$.

Therefore, the polar form of the complex number $\left(1-i\right)$ is $\left(1-i\right)=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$.

Simplify the expression ${\left(1-i\right)}^8$ by the use of De Moivre’s Theorem.

$\begin{array}{c}{\left(1-i\right)}^8={\left(\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\right)}^5\\ ={\left(\sqrt{2}\right)}^8\left[\cos \left(\left(\frac{7\pi }{4}\right)8\right)+i\sin \left(\left(\frac{7\pi }{4}\right)8\right)\right]\\ =2^4\left(\cos 14\pi +i\sin 14\pi \right)\\ =16\left(1+0i\right)\\ =16\end{array}$

Thus, the value of the complex number ${\left(1-\sqrt{3}i\right)}^5$ is $16$.