## Thirteen Power of (1+i)

If $i$ represents the imaginary unit, find real values for $a$ and $b$ such that $(1+i)^{13}=a+bi$.
Source: NCTM Mathematics Teacher, October 2006

Solution
$(1+i)^2=1+2i+i^2$
$=2i$
$(1+i)^{13}=((1+i)^2)^6(1+i)$
$=(2i)^6(1+i)$
$=64\:i^6(1+i)$
$=64(-1)(1+i)$
$=-64-64i$

Answer: $a=-64,b=-64$

Alternative solution
Write the complex number in polar form
$1+i=\sqrt{2}\left (\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\:i\right )$
$=\sqrt{2}\left (\text{cos} \dfrac{\pi}{4}+i\:\text{sin} \dfrac{\pi}{4}\right )$
Applying De Moivre’s theorem
$(1+i)^{13}=\left [\sqrt{2}\left (\text{cos}\dfrac{\pi}{4}+i\:\text{sin} \dfrac{\pi}{4}\right )\right ]^{13}$
$=(\sqrt{2})^{13}\left (\text{cos}\dfrac{13\pi}{4}+ i\:\text{sin}\dfrac{13\pi}{4}\right )$
$=64\sqrt{2}\left (\text{cos}\dfrac{5\pi}{4}+ i\:\text{sin}\dfrac{5\pi}{4}\right )$
$=64\sqrt{2}\left (-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\:i\right )$
$=-64-64i$