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?
 A: Note that on the circle $C$, given by $|z+i|=3$, we have $\overline{z+i}=\frac9{z+i}$.  Hence, on $C$
$$f(z)=\frac9{(z+i)^2}+i\frac{1}{z+i}$$
Then,
$$\begin{align}
\oint_{|z+i|=3}\frac{\overline{z}}{z+i}\,dz&=\color{red}{\oint_{|z+i|=3}\frac{9}{(z+i)^2}\,dz}+\color{blue}{i\oint_{|z+i|=3}\frac{1}{z+i}\,dz}\\\\
&=\color{red}{0}\color{blue}{-2\pi}
\end{align}$$


As an alternative approach, we will evaluate the integral using parameterization.  On $C$, $z=-i+3e^{i\theta}$ where $\theta \in [0,2\pi]$.  Furthermore, on $C$C we have $f(z)=\frac{\bar z}{z+i}=\frac{i+3e^{-i\theta}}{3e^{i\theta}}$ and $dz=i3e^{i\theta}\,d\theta$.  Hence,
$$\begin{align}
\oint_C f(z)\,dz&=\int_0^{2\pi} \left(-1+i3e^{-i\theta}\right)\,d\theta\\\\
&=-2\pi
\end{align}$$
as expected!
A: Observe that
$$
\frac{\overline{z}}{z+i}
=\frac{\overline{z}\cdot \overline{(z+i)}}{(z+i)\cdot \overline{(z+i)}}
=\frac{\overline{z}\cdot \overline{(z+i)}}{|z+i|^2}
=\frac{1}{9}\cdot \overline{z}\cdot \overline{(z+i)}
$$
Your parametrization of the path should be
$$
z(t)=\color{red}{-i}+3e^{2\pi i t},\quad [0,1]
$$
If your integral is $I$, then
\begin{align}
9I
&=\int_C \overline{z}\cdot \overline{(z+i)}\,dz\\
&=\int_0^1 (i+3e^{-2\pi i t})\cdot 3e^{-2\pi i t}\cdot 6\pi i\cdot  e^{2\pi it}
\,dt\\
&=\int_0^1 (i+3e^{-2\pi i t})\cdot 18\pi i\,dt=-18\pi
\end{align}
and thus $I=-2\pi$.
A: 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*}
