It should be solved with the Cauchy's Integral Formula. The solution given is $\displaystyle \frac{\pi}{2} + i\frac{\pi}{2}$ but I obtained $0$.
I did this: Let $C = \{|z-1|=2\}$
$$\displaystyle\int_C \frac{dz}{z^2-2i} = \int_C \frac{dz}{(z-(1+i))(z+(1+i))} $$ $$=\frac{1}{2+2i}\int_C \frac{dz}{z-(1+i)} - \frac{1}{2+2i}\int_C \frac{dz}{z+(1+i)}$$ With the Cauchy's formula, I get $$ \frac{1}{2+2i}\int_C \frac{dz}{z-(1+i)} = \frac{2\pi i }{2+2i}(f'(z_0))$$ If $z_0 = (1+i)$ and $f(z) = 1$, then $f'(1+i) = 0$. And analogously with $\displaystyle \frac{1}{2+2i}\int_C \frac{dz}{z+(1+i)}$.
Is there something wrong in my procedure?
Cauchy's integral formula is $$\int_C\frac{f(z)}{z-a}\,dz=2\pi i f(z)$$ when $C$ is a positively oriented contour, $a$ is inside $C$ and $f$ is holomorphic.
Letting $C$ be your contour, $a=1+i$ and $f(z)=1$ gives $$\int_C\frac{dz}{z-(1+i)}=2\pi i.$$
But $-1-i$ is outside $C$, so $$\int_C\frac{dz}{z+(1+i)}=0$$ by Cauchy's theorem.
So your original integral equals $$\frac{2\pi i}{2+2i}=\frac\pi 2(1+i).$$
But a simpler approach is to avoid partial fractions, and take $a=1+i$ and $f(z)=1/(z+1+i)$ in Cauchy's integral formula.