Chapter 8, Problem 36RE

To determine

De Moiré’s theorem and find the indicated power. ${\left(1+i\right)}^8$

Answer to Problem 36RE

The indicated power of ${\left(1+i\right)}^8$ is $16$ .

Explanation of Solution

Given:

  ${\left(1+i\right)}^8$

Concept Used:

De Moiré’s theorem that state for any complex number.

  ${\left(\cos \theta +i\sin \theta \right)}^n=\left(\cos n\theta +i\sin n\theta \right)$

Calculations:

First convert the complex number in the polar form the equation is,

  $a+ib=r\left(\cos \theta +\sin \theta \right)$

Now compare real and imaginary parts,

  $\begin{array}{l}a=r\cos \theta ,\\ b=r\sin \theta \end{array}$

Now squaring and adding real and imaginary parts now find the value of $r$ ,

  $\begin{array}{l}{r}^2=a^2+b^2\\ r=\sqrt{a^2+b^2}\end{array}$

Angle can be found by dividing imaginary by real part of the complex number and quadrant can be found using sign of real $x$ and imaginary $y$ part, so the angle can be found accordingly.

  $\tan \theta =\frac{b}{a};\theta {\tan }^{-1}\left(\frac{b}{a}\right)$

So, the given complex number is ${\left(1+i\right)}^8$

Now by comparing the $a+ib$ , the simplification is $a=1,b=1$ .

Now calculate the value of $r$ ,

  $r=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$

Now finding the value of $\theta$ ,

  $\theta ={\tan }^{-1}\left(\frac{1}{1}\right)=\frac{\pi }{4};$ as $x:+ve,y:-ve\text{so}1st\text{quaderant}\text{.}$

Now after conversion to polar form,

  ${\left(1+i\right)}^8={\left(\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right)}^8$

Now by applying the de moivres formula theorem,

  ${\left(1-\sqrt{3}i\right)}^4=2^4\left(\cos \frac{4\times 5\pi }{3}+i\sin \frac{4\times 5\pi }{3}\right)$

Upon simplifying,

  $\begin{array}{l}{\left(1+i\right)}^8={\left(\sqrt{2}\right)}^8{\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}^8\\ {\left(1+i\right)}^8={\left(\sqrt{2}\right)}^8\left(\cos \frac{8\pi }{4}+i\sin \frac{8\pi }{4}\right)\\ {\left(1+i\right)}^8={\left(\sqrt{2}\right)}^8\left(\cos 2\pi +i\sin 2\pi \right)\\ {\left(1+i\right)}^8=2^4\left(\cos 2\pi \right)\\ {\left(1+i\right)}^8=16\end{array}$

Conclusion:

Hence, the indicated power of ${\left(1+i\right)}^8$ is $16$ .