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I want to evaluate $$\oint_\gamma \sqrt z \, dz \, ;\quad\gamma=\{z:|z|=1\}$$ and I know I can do so by parameterising the curve and plugging in the relevant formulas. However I tried to evaluate this integral using $z^{3/2}$ as the primitive to $\sqrt z$ and could not reach the same answer.

If I define $\sqrt z:= e^{\frac{1}{2}Logz}$ and split $\gamma$ into two curves $\gamma_1, \gamma_2$ representing the half circles from $-i$ to $i$ and from $i$ to $-i$ with a counter-clockwise orientation then my method to evaluate the integral was as follows:

$\gamma_1=\{z:z=re^{i\theta}, \, \theta \in [-\frac{\pi}{2},\frac{\pi}{2}]\} , \quad\gamma_2=\{z:z=re^{i\theta}, \, \theta \in [\frac{\pi}{2},\frac{3\pi}{2}]\}$

Also, along $\gamma_1$ I define $z^{3/2}=e^{\frac{3}{2}\text{Log}z}$ with Log$z$ having branch cut discontinuity on the negative real line and define a new Log$_0z$ with a similar definition to Log$z$ but with branch cut discontinuity along the positive real line $[0,\infty)$ and define $z^{3/2}=e^{\frac{3}{2}\text{Log}_0z}$ along $\gamma_2$.

\begin{align} \oint_\gamma \sqrt z \, dz=\int_{\gamma_1} \sqrt z \, dz+\int_{\gamma_2} \sqrt z \, dz&=\bigg[\frac{2}{3}z^{3/2}\bigg]^i_{-i}+\bigg[\frac{2}{3}z^{3/2}\bigg]^{-i}_{i} \\&=\bigg[\frac{2}{3}e^{\frac{3}{2}\text{Log}z}\bigg]^i_{-i}+\bigg[\frac{2}{3}e^{\frac{3}{2}\text{Log}_0z}\bigg]^{-i}_{i} \end{align}

and now Log$(i)=\text{ln}(1)+i\text{Arg}(i)=\frac{\pi i}{2}$ and similarly Log$(-i)=-\frac{\pi i}{2}$, $\,$ Log$_0(i)=\frac{\pi i}{2}, \,$Log$_0(-i)=\frac{3\pi i}{2}$ with $\text{Arg}_0 \in (0, 2\pi]$ and now doing the calculation I get something different to parameterising the curve and integrating as such. Any insight as to where I could have gone wrong?

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    $\begingroup$ This integral does not have a well defined value, not even up to a factor $\pm1$. $\endgroup$ – Christian Blatter Feb 23 '18 at 15:00
  • $\begingroup$ But taking a parameterisation of $\gamma$ it does? $\endgroup$ – user258521 Feb 23 '18 at 15:22
  • $\begingroup$ you have to specified the branch of the function rather than taking a parameterisation $\endgroup$ – mastrok Feb 23 '18 at 15:46
  • $\begingroup$ Parameterise the curve by $z(t)=e^{it}$ for $t \in (-\pi,\pi]$ and then we get a value for the integral $\endgroup$ – user258521 Feb 23 '18 at 17:09

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