Calculate $\int_{\gamma}{dz \over \sqrt{z}}$, where $\gamma$ is a section connecting points $z=4$ and $z=4i$, and $\sqrt z$ is the side of a function, where $\sqrt 1 = 1$
The line that I found is $y=i(4-x)$, now assuming $z = e^{x+iy}$, and $\sqrt z = e^{{x+iy + 2\pi k} \over 2}$ and $\sqrt 1 = e^{x+2\pi k \over 2}$ seems doesn't give me anything, I have solve this type of problems where $|z| = a$ and function is a circle, but with section I don't get it. Any helps are welcome.
On
In $\mathbb{C}\setminus(-\infty,0]$, you have the main branch of the square root of $z$, $\sqrt z$. A primitive of $\frac1{\sqrt z}$ is $2\sqrt z$. So, your integral is equal to$$2\left(\sqrt{4i}-\sqrt4\right)=2\left(\sqrt2(1+i)-2\right).$$
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\gamma}{\dd z \over \root{z}} & = -\int_{4}^{0}{\ic\,\dd y \over \root{\ic y}} - \int_{0}^{4}{\dd x \over \root{x}} = \pars{\expo{\ic\pi/4} - 1}\int_{0}^{4}{\dd x \over \root{x}} \\[5mm] & = \bracks{\pars{{\root{2} \over 2} - 1} + {\root{2} \over 2}\,\ic}4 = \bbx{2\bracks{\pars{\root{2} - 2} + \root{2}\,\ic}} \end{align}
On $\gamma$, $z = 4e^{it}$, $t \in [0,\dfrac\pi2]$. $dz = 4ie^{it} \, dt$
\begin{align} \int_\gamma \frac{dz}{\sqrt{z}} &= \int_0^{\pi/2} \frac{4ie^{it} \, dt}{2e^{it/2}} \\ &= \int_0^{\pi/2} 2i e^{it/2} \, dt \\ &= [4e^{it/2}]_0^{\pi/2} \\ &= 4(e^{i\pi/4}-1) \\ &= 4(\frac{1+i}{\sqrt2} - 1) \\ &= 2 \sqrt2 [(1-\sqrt2) + i] \end{align}