Line integral of complex expression

Question

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.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{Z = 1 + \ic + \pars{2 + \ic}\mu\quad}$ 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}

Answer 2

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 $$ \int_C \operatorname{Re}(z)\,dz = \int_{t=0}^{t=1} x\, (dx+i\,dy) = \int_0^1 ((1-t)(1)+t(3)) \, (2\,dt + i\,dt) $$ $$ =\int_0^1 ((1-t)(1)+t(3))\, 2\,dt + i\int_0^1 ((1-t)(1)+t(3))\,dt. $$