How to calculate $\int_{C}{}\frac{\bar{z}}{z+i} dz$ if $C$ is a circle: $|z+i|=3$?

Question

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?

Answer 1

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 $I=-2\pi$.

Answer 2

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*}

Answer 3

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,

$$\begin{align} \oint_{|z+i|=3}\frac{\overline{z}}{z+i}\,dz&=\color{red}{\oint_{|z+i|=3}\frac{9}{(z+i)^2}\,dz}+\color{blue}{i\oint_{|z+i|=3}\frac{1}{z+i}\,dz}\\\\ &=\color{red}{0}\color{blue}{-2\pi} \end{align}$$

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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!