Integration of a part of the circle

Question

I need to find a proper solution for the integral. I could not do anything about it. If you try to help me that would be great. Thank you

$\displaystyle \int_\gamma$ ( $i\overline z$+$z^2)\mathrm{d}z =\ ?$

$\gamma$ is the part of the circle and $\lvert z\rvert = 2$ ; arg(z) $\in$ $[\pi/2,\pi]$

Answer 1

A suitable parametrization of $\gamma$ would be $\gamma(t)=2e^{\mathrm it},~t\in[\pi/2,\pi]$. Then just plug it into the definition. If $\gamma$ is a paremtrization on $[a,b]$, then

$$\begin{align*} \int_\gamma f(z)\mathrm dz&:=\int_a^b \gamma'(t)f(\gamma(t))\mathrm dt\\ \int_\gamma (\mathrm i\bar z+z^2)\mathrm dz&=\int_{\pi/2}^\pi\underbrace{2\mathrm ie^{\mathrm it}}_{\gamma'}(\mathrm i\overline{2e^{\mathrm it}}+\left(2e^{\mathrm it}\right)^2)\mathrm dt\\ &=\int_{\pi/2}^\pi 2\mathrm i e^{\mathrm it}(2\mathrm ie^{-\mathrm it}+4e^{2\mathrm it})\mathrm dt\\ &=\int_{\pi/2}^\pi 4\left(-e^0+2\mathrm ie^{3\mathrm it}\right)\mathrm dt \end{align*}$$

You just need to use the rules for the complex exponential function: $\overline{\exp(z)}=\exp(\bar z)$ and then the usual rules known from the reals. And from here, integrating should be doable.

Answer 2

You are on the circle $|z|=2$ therefore

$z\overline z=|z|^2=4,\overline z=4/z$.

So you may render the integrand as

$i\overline z +z^2=(4i/z)+z^2$

and integrate the separate terms, thus

$\int (i\overline z+z^2) dz = \int ((4i/z)+z^2) dz = 4i\ln z+(z^3)/3+C$.

To handle the logarithm: Assume the argument varies continuously from $\pi/2$ to $\pi$, that is you do not cross the branch cut.

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} &\bbox[5px,#ffd]{\int_{z\ \in\ 2\expo{\large\ic\pars{\pi/2,\pi}}}\pars{\ic\overline{z} + z} \,\dd z} = \int_{\pi/2}^{\pi}\bracks{\ic\pars{2\expo{-\ic\theta}} + 2\expo{\ic\theta}}\,2\expo{\ic\theta}\ic\,\dd\theta \\[5mm] = &\ \int_{\pi/2}^{\pi}\pars{-4 + 4\ic\expo{2\ic\theta}}\dd\theta = \bbx{4 - 2\pi} \\ &\ \end{align}