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We say a function $f$ has a continuous root, if there exists a continuous function $g$ such that $g^{2}(z) = f(z)$ for all $z\in\Omega$, where $\Omega$ is some open domain.

Let $f:B_r(0)\to\mathbb{C}$ be analytic, $f(0)=0$. WLOG, we can assume $f(z)\neq 0$ for $z\in\Omega, z\neq 0$. Then there exists $k\in\mathbb{N}$, $h(z)$ analytic such that $f(z) = z^kh(z)$, $h(z)\neq 0$ for all $z\in\Omega$. Suppose $f$ has a continuous root, that is there exists $g:\Omega\to\mathbb{C}$ continuous such that $g^2(z) = f(z)$. I want to show that this means that $k$ is necessarily even.

By the given, we have that $$ g^2(z) = z^kh(z) $$ I should let you know now that we have not defined a square root function $\sqrt{z}$ in the course, so to me there is no meaning to $\sqrt{h(z)}$, or any non integer power in that manner.

But this is where I am stuck. I am trying to show that if $k$ is odd there $g$ cannot be continuous at the real axis, but if $z = r\in\mathbb{R}$, then $$ g^2(r) = r^{2m+1}h(z) $$

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  • $\begingroup$ Your $g^2(z)$ means $g(z)^2$, not $g(g(z))$, complying with your further considerations? $\endgroup$ – LutzL Dec 20 '17 at 17:21
  • $\begingroup$ Yes, $g^2 = g\cdot g$ $\endgroup$ – Joshhh Dec 20 '17 at 17:22
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Well, whether you've proved it or not, there is a continuous square root of $z$ in any disk $D(p,r)$, if $0<r<|p|$.

Hence $h$ has a continuous square near the origin, so if $f$ has a continuous square root near the origin then so does $z^k$.

In polar coordinates $z^k=r^ke^{ik\theta}$. It follows that the only square roots of $z^k$ have the form $\pm r^{k/2}e^{ik\theta/2}$, and this is not continuous near the origin unless $k$ is even.

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