## Calculus 8th Edition

$16$
We have: $z=1-i=1+-1*i$ 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 4th quadrant, we have: $\displaystyle \theta=\frac{7\pi}{4}$ To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$: $z=\displaystyle \sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})$ We apply the exponent to the polar form of the function and use De Moivre's Theorem: $(1-i)^{8}=[\displaystyle \sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})]^{8}=2^{4}(\cos\frac{8*7\pi}{4}+i\sin\frac{8* 7\pi}{4})=16(\cos 14\pi+i\sin \mathrm{l}4\pi)=16(1+0i)=16$