Calculate integral $$\oint_{\gamma} \frac{z+1}{z^4+2iz^3} dz$$ where $\gamma$ is parameterization of circle $B(0,1)$ along one positive rotation.
I did something like this with Cauchy. \begin{align} \oint_{\gamma} \frac{z+1}{z^4+2iz^3}dz&=\oint_{\gamma} \frac{z+1}{z^3(z+2i)} =\oint_{\gamma} \frac{\dfrac{z+1}{z+2i}}{(z-0)^{2+1}}dz \\ &=\frac{2i\pi}{n!}f^{''}(z_0) \\ &=\frac{2i \pi}{2!}\left(\frac{2-4i}{(0+2i)^3}\right) \\ &=\frac 12 \pi i-\frac 14 \pi \end{align}
with this one I'm not that sure
Calculate integral $$\oint_{\gamma} \frac{z^3+3}{z(z-i)^2}dz $$ Where $\gamma$ is 8-like curve where the rotation is positive around $(0,i)$ and negative around $(0,0)$.
Isn't the integral zero over $\oint_{\gamma_1}(\text{positive part})$ + $\oint_{\gamma_2}(\text{negative part})=0$. So is it two integrals were $\oint_{\gamma_2}=-\oint_{\gamma_1}$.
And the integral would be \begin{align}\oint_{\gamma_1}&=\frac{\dfrac{z^3+3}{z}}{(z-i)^2}dz \\ &=\frac{2\pi i}{1!}\left( \frac{2(-i)^3 -3}{(-i)^2}\right) \\ &=2\pi i (3-2i)\\ &=4\pi+6\pi i \\ \oint_{\gamma_2}&= -4\pi-6\pi i \end{align}
Did I get these question correct? I'm not 100% on the second question.