a) $(-1+i)^{-i}$ so I know that the answer is $9.92-3.58i$. My track getting there is off.
I know that $x=-1$ and $y=1$, so $r = \sqrt{2}$, also that $\displaystyle \theta=-\frac{pi}{4}$.
I've plugged them into de Moivre's but my final answer does not match the right answer. Help?
b) $(-1)^{1/pi}$
Let $-1+i=r(\cos\theta+i\sin\theta)=r\cdot e^{i\theta}$ (Using Euler's Identity)
Equating the real & the imaginary parts
$-1=r\cos\theta, 1=r\sin\theta$
Squaring and adding we get $r^2=2\implies r=\sqrt2$
$\implies \cos\theta=-\frac1{\sqrt2}<0, \sin\theta=\frac1{\sqrt2}>0 $
From the definition of $\arctan\frac yx,\theta= \pi+\arctan\frac1{(-1)}=\pi-\frac{\pi}4 =\frac{3\pi}4$ as the principal value of arctan lies $\in\left[-\frac\pi2, \frac\pi2\right]$
So, $$(-1+i)^i=e^{i\ln(-1+i)}=e^{i\ln(\sqrt 2e^{\frac{3\pi i}4})}$$
$$=e^{i(\frac12\ln2+\frac{3\pi i}4)}$$
$$=e^{-\frac{3\pi}4}\cdot e^{\left(i\frac{\ln 2}2\right)}$$
$$=e^{-\frac{3\pi}4}\left(\cos \frac{\ln 2}2+i\sin \frac{\ln 2}2 \right)$$