# 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.

• you can deform the segment to a piece of a circle while keeping its ends fixed without passing over singular points of the holomorphic function $\frac{1}{\sqrt{z}}$, so the integral does not change. – Orest Bucicovschi Apr 6 '18 at 22:49

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}

• How is that $z = 4e^{it}$ ? – Spike Bughdaryan Apr 6 '18 at 22:27
• @SpikeBughdaryan $z = 4e^{it} = 4(\cos t + i\sin t)$ represents the quarter of the circle $|z|=4$ in the first quadrant. – GNUSupporter 8964民主女神 地下教會 Apr 6 '18 at 22:32
• How can we take a circle? Do we have the independence of the road? – Spike Bughdaryan Apr 6 '18 at 22:40
• @SpikeBughdaryan Yes, as $\frac{1}{\sqrt{z}}$ is analytic on $\mathbb{C}\setminus(-\infty,0]$. So by Cauchy's Theorem, the contour integral along a curve inside that region is zero. – GNUSupporter 8964民主女神 地下教會 Apr 6 '18 at 22:47

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$$2\left(\sqrt{4i}-\sqrt4\right)=2\left(\sqrt2(1+i)-2\right).$$

• Where did You used $\sqrt 1=1$ ? – Spike Bughdaryan Apr 6 '18 at 22:23
• I used the main branch of the square root, for which the square root of $1$ is $1$. – José Carlos Santos Apr 6 '18 at 22:24

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}