Answer
$-1024$
Work Step by Step
Given: $z=1+i=1+1*i$
To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$:
$r=\sqrt{1^2+1^2}=\sqrt{2}$
To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$:
$\displaystyle \tan\theta=\frac{1}{1}=1 $
And since (1,1) is in the 1st quadrant:
$\displaystyle \theta=\frac{\pi}{4} $
To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$:
$z=\displaystyle \sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Apply the exponent to the polar form of the function and use De Moivre's Theorem:
$(1+i)^{20}=[\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}
{4})]^{20}$
$=(2^{1/2})^{20}(\cos\frac{20\pi}{4}+i\sin\frac{20\pi}{4})$
$=2^{10}(\cos 5\pi+i\sin 5\pi)
=2^{10}(-1+i\times 0)$
$=-2^{10}$
$=-1024$
