How can I compute the integral $\int_{\psi(0,1)}\frac{z+z^*}{z-1/2}$? I am working out some exercises that came with some notes on complex analysis. Forgive me if the answer is obvious, it's just not my day today. The tools I can use are:

*

*Cauchy's Integral Theorem

*Cauchy's Integral Formula (also for derivatives)

*Deformation Theorem

*A bunch of other theorems but they all rely on the integrand being holomorphic inside and/or the region enclosed by the contours.

*I can't use the Residue Theorem because it's not covered in the notes

$$\int_{\psi(0,1)} \frac{Re(z)}{z-1/2}\text{ }dz$$
By $\psi_{(0,1)}$ we understand the set of points $\{z:|z| = 1\}$.
Here's what I've tried:

*

*The first time I saw it, I tried to use Cauchy's Integral Formula. No good, because $Re(z)$ is not holomorphic anywhere.

*I had the idea to rewrite $Re(z)$ as: $$\frac{z+z^*}{2}$$ and split the integral. This allows me to work out the first half using Cauchy's, but not the second half.

*I tried rewriting the numerator as $z - i Im(z)$ but this looks about the same problem shifted on $Im(z)$ instead.

*I tried rewriting the integral in its explicit parameterisation. It didn't end up simplifying much. I arrived at:$$\int_0^{2\pi}\frac{i(\cos^2 (t) + \cos(t)\sin(t))}{e^{it}-\frac{1}{2}}dt$$

*I am not very confident when it comes to variable substitution in complex analysis. I suspect this might make the integral much easier to work it. I understand from the notes that a reparameterisation must be a 'continuously differentiable bijective function'.

The answer is listed as $\frac{i\pi}{2}$.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{\verts{z}\ =\ 1}{\Re\pars{z} \over z - 1/2}\,\dd z & =
\int_{\verts{z}\ =\ 1}{\pars{z + \overline{z}}/2 \over z - 1/2}\,\dd z =
{1 \over 2}\int_{\verts{z}\ =\ 1}{z + \pars{z\overline{z}}/z \over z - 1/2}\,\dd z
\\[5mm] & =
{1 \over 2}\int_{\verts{z}\ =\ 1}{z^{2}  + 1 \over z\pars{z - 1/2}}\,\dd z =
{1 \over 2}\, 2\pi\ic
\pars{{0^{2} + 1 \over 0 - 1/2} + {\pars{1/2}^{2} + 1 \over 1/2}}
\\[5mm] & =
\pi\ic\pars{-2 + {5 \over 2}} = \bbx{\large{\pi \over 2}\,\ic} \\ &
\end{align}
