How to calculate $\int_{C}{}\frac{\bar{z}}{z+i} dz$ if $C$ is a circle: $|z+i|=3$? The task:
$$
\text{Calculate } \int_{C}{}f(z)dz\text{, where } f(z)=\frac{\bar{z}}{z+i}\text{, and } C \text{ is a circle } |z+i|=3\text{.}
$$
Finding the circle's center and radius:
$$
|z+i|=|x+yi+i|=|x+(y+1)i|=3
\\
x^2+(y+1)^2=3^2
$$
Parametrizing the circle:
$$
z(t)=i+3e^{-2\pi i t}
$$
Now I need to calculate this integral:
$$
\int_{0}^{1}{f(z(t))z'(t)dt}=
2\pi\int_{0}^{1}{
\frac{1-3ie^{-2\pi i t}}{e^{4 \pi i t}}dt
}
$$
Unfortunately I calculated this integral, and it's equal to $0$. Is this correct? I don't think so. Where did I go wrong? Maybe I made a mistake when calculating the integral - what would be the best way to calculate it?

You cannot directly use the Residue Theorem in this case since $\frac{\overline{z}}{z+i}$ is not holomorphic, so your approach is entirely reasonable. Your calculation of the integral is correct since
$$\int_{0}^{1}e^{2k\pi ti}dt=\left.\frac{1}{2\pi ki}e^{2k\pi kti}\right|_0^{1}=\frac{1}{2\pi ki}(1-1)=0$$
for any integer $k$, and so
\begin{align*}
2\pi\int_{0}^{1}\frac{1-3ie^{-2\pi it}}{e^{4\pi i t}}dt&=2\pi\int_{0}^{1}e^{-4\pi i t}dt-6\pi i\int_{0}^{1}e^{-6\pi it}dt\\
&=2\pi(0)-6\pi i(0)\\
&=0
\end{align*}
