Solution. (a)

$$

\begin{aligned}

\frac{i}{1-i}+\frac{1-i}{i} &=\frac{i^{2}+(1-i)^{2}}{(1-i) i} \\

&=\frac{-1-2 i}{1-i} \cdot \frac{1-i}{1-i} \\

&=\frac{-1+i-2 i-2}{2} \\

&=\frac{-3-i}{2}=-\frac{3}{2}-\frac{i}{2}

\end{aligned}

$$

(b) We have that

$$

-8 i=2^{3} \exp \left(\frac{-i \pi}{2}\right)

$$

Thus the cube roots are

$$

2 \exp \left(\frac{-i \pi}{6}\right), \quad 2 \exp \left(\frac{i \pi}{2}\right) \text { and } 2 \exp \left(\frac{7 i \pi}{6}\right)

$$

That is

$$

\sqrt{3}-i, \quad 2, \quad-\sqrt{3}-i

$$

\begin{array}{l}

\text { (c) }\\

\begin{aligned}

\left(\frac{i+1}{\sqrt{2}}\right)^{1337} &=\left(\exp \frac{i \pi}{4}\right)^{1337} \\

&=\exp \frac{1337 \pi i}{4} \\

&=\exp \left(167 \cdot 2 \pi i+\frac{\pi}{4} i\right) \\

&=\exp \frac{\pi}{4} i=\frac{1+i}{\sqrt{2}}

\end{aligned}

\end{array}