# Integration of $z^{1/2}$ along a contour winding the origin twice, without introducing Riemann surface

If we introduce a Riemann surface, it is easy to show that the integral of $z^{1/2}$ along a contour winding the origin twice is zero. An anti-derivative exists everywhere, so the integral depends only on the end points of the contour. Thus, every closed contour in the surface gives zero. Is there any way to ubderstand this result only with branchcuts and without introducing multi-valued functions or Riemann surfaces?

• What is your definition of $z^{1/2}$ that you want to integrate? – Christian Blatter Aug 4 '13 at 11:38
• @Blatter, it is just usual $\sqrt{r}e^{i \theta/2}$. The domains are usual Riemann surface of it with two sheets in one case and the complex plane with a branchcut from the origin to infinity in the other case. – asofas Aug 4 '13 at 11:49

Cut the complex plane along the negative real axis. On the resulting region $\Omega$ the function $${\rm Log}(z):=\log |z|+ i{\rm Arg}(z)$$ is well defined and analytic. It follows that $$f(z):=\exp\left({1\over 2}{\rm Log}(z)\right)$$ is an analytic square root of $z$ on $\Omega$, and furthermore $f$ has an analytic primitive $$F(z):={2\over 3}\exp\left({3\over 2}{\rm Log}(z)\right)$$ on $\Omega$.
Assume now that your cycle $\gamma$ leaves the cut downwards at $-b$, goes around the origin counterclockwise and hits the cut again from above at $-a$. It then makes a second tour around the origin and finally returns to $-b$. What is essential here is the following: When we are leaving the point $-a$ for the second tour we must not use the above $f$ as integrand, but $-f$ in order to make the integrand analytic along the whole of $\gamma$. Note that when arriving at $-a$ from above the limiting value of $f$ is $\sqrt{a}e^{i\pi/2}=i\sqrt{a}$, whereas when approaching $-a$ from below the limiting value of $f$ is $\sqrt{a}e^{-i\pi/2}=-i \sqrt{a}$.
Using analogous care with $F$ when approaching $-a$, $-b$ from above, resp. from below, you will see that there is a lot of cancellation, so that in the end we obtain $$\int\nolimits_\gamma z^{1/2}\ dz=0\ ,$$ under the condition that $z^{1/2}$ is interpreted properly along $\gamma$.