Line integral of complex expression How can we integrate expressions like these $\int_C \operatorname{Re}(Z) \, dZ$ where $C$ is the  shortest path joining the points $1+i$ and $3+2i$.
The $\operatorname{Re}(Z)$ in the expression is what confusing me.
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{Z = 1 + \ic + \pars{2 + \ic}\mu\quad}$ with $\ds{\quad 0 \leq \mu \leq 1}$.
\begin{align}
\color{#00f}{\large\int_{C}\Re\pars{Z}\,\dd Z}&=
\int_{0}^{1}\pars{1 + 2\mu}\,\bracks{\pars{2 + \ic}\,\dd\mu}
=\pars{2 + \ic}\int_{0}^{1}\pars{1 + 2\mu}\,\dd\mu
=\left.\pars{2 + \ic}\pars{\mu + \mu^{2}}\right\vert_{0}^{1}
\\[3mm]&=\color{#00f}{\large 4 + 2\ic}
\end{align}
A: As $t$ goes from $0$ to $1$, then $(1-t)(1+i) + t(3+2i)$ goes from $1+i$ to $3+2i$ along a straight line.  So let $z=x+iy=(1-t)(1+i) + t(3+2i)$, so that $\operatorname{Re}(z) = x$.  Then
$$
\int_C \operatorname{Re}(z)\,dz = \int_{t=0}^{t=1} x\, (dx+i\,dy) = \int_0^1 ((1-t)(1)+t(3)) \, (2\,dt + i\,dt)
$$
$$
=\int_0^1 ((1-t)(1)+t(3))\, 2\,dt + i\int_0^1 ((1-t)(1)+t(3))\,dt.
$$
