Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$

Question

In my test on complex analysis I encountered following problem:

Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$

So first I observed that function $z\text{Im}(z)$ is not holomorphic at least on real axis. Therefore we have to intgrate using parametrization.

First, let's change variable $w = z - \frac{1}{3}$. So we got $\oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w+\frac{1}{3})\text{d}w = \oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w)\text{d}w = \frac{1}{2i}\oint\limits_{|w|=3} (w+\frac{1}{3}) (w-\bar w)\text{d}w$.

Then by letting $w=3e^{i \phi}$ we transform integral to the form $\frac{1}{2}\int\limits_{0}^{2\pi}(3e^{i \phi}+\frac{1}{3})(3e^{i \phi}-3e^{-i \phi})ie^{i \phi}\text{d}\phi = -\frac{1}{2}\int\limits_{0}^{2\pi}\text{d}\phi=-\pi$.

Is my reasoning correct? I don't quite sure about change of variable I made since function is not holomorphic at real axis. Is there any other way how this integral can be evaluated? Thanks!

Answer 1

Note that since $\text{Im}(z)=\frac1{2i}(z-\bar z)$, that

$$z\text{Im}(z)=\frac1{2i}(z^2-|z|^2)$$

Since $z^2$ is analytic, we have

$$\begin{align} \oint_{|z-\frac13 |=3}z\text{Im}(z)\,dz&=\frac i2\oint_{|z-\frac13 |=3}|z|^2\,dz\\\\ &=-\frac {3}2 \int_0^{2\pi} \left|\frac13 +3e^{i\phi}\right|^2 e^{i\phi}\,d\phi\\\\ &=-\frac {3}2 \int_0^{2\pi} \left(\frac{10}9 +e^{i\phi}+e^{-i\phi}\right)e^{i\phi}\,d\phi\\\\ &=-3\pi \end{align}$$

Answer 2

You can also use Stokes'theorem. Since $z\mathrm{Im}(z)=\frac{1}{2i}(z^2-z\overline{z})$ and $dx\wedge dy=\frac{1}{2i}d\overline{z}\wedge dz$ we have $$ \oint_{|z-\frac{1}{3}|=3}\frac{z^2-z\overline{z}}{2i}dz=\iint_{|z-\frac{1}{3}|\le3}-z\frac{d\overline{z}\wedge dz}{2i}=\iint_{(x-\frac{1}{3})^2+y^2\le 9}(-x-iy)dx\wedge dy=\\ =-\iint_{(x-\frac{1}{3})^2+y^2\le 9}xdx\wedge dy-i\iint_{(x-\frac{1}{3})^2+y^2\le 9}ydx\wedge dy=\\ =-Average_D(x)\times Volume(D)-iAverage_D(y)\times Volume(D)=\\ =-\frac{1}{3}\times 9\pi-i\cdot 0\times 9\pi=-3\pi $$ where $D=\{(x,y)\in\mathbb{R}|(x-\frac{1}{3})^2+y^2\le 9\}$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \oint_{\verts{z - 1/3}\ =\ 3}z\,\Im\pars{z}\,\dd z & = \int_{-\pi}^{\pi}\pars{{1 \over 3} + 3\expo{\ic\theta}} \Im\pars{{1 \over 3} + 3\expo{\ic\theta}}3\expo{\ic\theta}\ic\,\dd\theta \\[5mm] & = 3\ic\int_{-\pi}^{\pi}\pars{{1 \over 3}\expo{\ic\theta} + 3\expo{2\ic\theta}} \bracks{3\sin\pars{\theta}}\dd\theta \\[5mm] & = 3\ic\int_{-\pi}^{\pi}\bracks{\sin^{2}\pars{\theta}\ic + 9\sin\pars{2\theta}\sin\pars{\theta}\ic}\dd\theta \\[5mm] & = -6\int_{0}^{\pi}\bracks{{\color{red}{1} - \cos\pars{2\theta} \over \color{red}{2}} + 9\,{\cos\pars{\theta} - \cos\pars{3\theta} \over 2}}\dd\theta \\[5mm] & = -6\int_{0}^{\pi}{1 \over 2}\,\dd\theta = \bbx{-3\pi} \end{align}