Find complex integral $\int_{\gamma}{dz \over \sqrt{z}}$ 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.
 A: 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$$2\left(\sqrt{4i}-\sqrt4\right)=2\left(\sqrt2(1+i)-2\right).$$
A: 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}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\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}
