Evaluating integral on 2 different ways Evaluate on two different way the following line integral.
$\int_{\partial K_1(0)}^{} \! \sqrt{z}  \, dz $
So I am not sure which two ways. I was thinking maybe something with residue theorem but this function has no poles.
P.S. I am not sure how to explain what K here means, it is connected with circle (german) but I'm not sure what excatly it means.
 A: Principal branch
$$
\sqrt{z} = e^{\frac{1}{2}\ln z}
$$
where the principal branch of $\ln z$ is defined as $\mathbb{C} \backslash[0,\infty)$.
The square root function is plotted below. The plot suggests the contour integral will be imaginary.

@Errol.Y advocates for clarification. Consider the plots in 3D. Imagine that you are standing at the point $-1+i 0$ in the complex plane. Move along the contour clockwise and keep track of your altitude changes. The real part of the function returns to the starting value; there is no net change. However, the complex part of the function records a drop of $4/3$ units.

Method 1
Parameterization of the contour $C$, the unit circle from $\theta=-\pi$ to $\theta=\pi$:
$$
z = e^{i \theta}, dz = i e^{i \theta} d\theta
$$
The parameterized integral is
$$
\int_{C} \sqrt{z} dz = \int_{-\pi}^{\pi} e^{i \frac{\theta}{2}}  i e^{i \theta} d\theta= \frac{2}{3} \left( e^{i \frac{3\pi}{2}} - e^{-i \frac{3\pi}{2}}\right) = - \frac{4}{3}i
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\root{z} = \verts{z}^{1/2}\exp\pars{{1 \over 2}\,\mrm{arg}\pars{z}\ic}\,;
\qquad z \not= 0\,,\quad -\pi < \,\mrm{arg}\pars{z} < \pi}$



*

*The integration is performed along a key-hole contour which 'takes   care' of the $\ds{\root{z}}$ branch cut.

*The contribution from an indented path
$\ds{\pars{~\mbox{a semi-circumference of radius}\ \epsilon~}}$ 'around'  the origin  obviously vanishes out in the $\ds{\epsilon \to 0^{+}}$-limit.  



Since there isn't any pole 'inside' the contour, the whole integration is reduced to:
\begin{align}
\oint_{\partial\mrm{K}_{1}\pars{0}}\root{z}\,\dd z & =
-\int_{-1}^{0}\root{-x}\expo{\ic\pi/2}\,\dd x -
\int_{0}^{-1}\root{-x}\expo{-\ic\pi/2}\,\dd x
\\[5mm] & =
-\ic\int_{0}^{1}\root{x}\,\dd x +
\pars{-\ic}\int_{0}^{1}\root{x}\,\dd x = \bbx{\ds{-\,{4 \over 3}\,\ic}}
\end{align}
