# Holomorphic n-th roots

Suppose we have a simply connected domain in the complex plane containing 0. I would like to show that there is no holomorphic n'th root function defined on the domain.

Using some student logic, I feel that I have to use the log function and its properties. Indeed, log is not defined at 0 so it seems I could come up with a contradiction if I could relate a hypothetical n'th root function on the domain to log.

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Simple connectivity is irrelevant for this problem; you could just show it for a disk centered at $0$, because such a disk is contained in any domain containing $0$. Since you know how taking roots treats modulus, it is enough to show that there is no continuous $n^\text{th}$ root on the circle $|z|=1$. – Jonas Meyer Feb 7 '11 at 19:13
To say it a bit differently, any domain containing the origin will contain a ball B(0,e+). Then $e^1/2$=$((-e)^1/2$. – gary May 14 '11 at 23:56
Sorry, I meant, in a ball B(0,e) contained in the region, consider the images of , e/2 and -e/2 under $z^1/2$ . – gary May 15 '11 at 16:29

Pick a circle around the origin that lies within the domain and move around it once. If you have a holomorphic $n$-th root function, you get the same function value again when you get back to where you started. But that can't be, since the argument (angle in polar coordinates) of your point and the argument of the function value must vary continuously with each other, with the argument of the function value increasing at one n-th of the rate of the argument of your point. Since the argument of your point has increased by $2\pi$, the argument of the function value must have increased by $2\pi/n$, so you can't get the same function value again. The contradiction shows that there can be no such function.
Here's a simple proof by contradiction: Cauchy's theorem tells us that the integral of a holomorphic function over a closed curve in a simply-connected domain vanishes. Show that this is not the case for $$\oint_C z^{\frac{1}{n}} \, dz$$ where C is the unit circle.