Integration of a part of the circle 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) $\in$ $[\pi/2,\pi]$
 A: 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.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
