# Is $f(z)=\sqrt z$ differentiable in the complex plane?

I am wondering if someone could help me with following complex analysis question:

Is $f(z)=\sqrt z$ differentiable in the complex plane?

I think the answer will be everywhere but for $\theta=-\pi$ but not sure if it is the correct answer, and if it is for what reason.

• How is $\sqrt{z}$ defined and what is its domain? – Dimitris May 16 '16 at 5:11
• – Watson Mar 22 '18 at 12:25

You can't even define $\sqrt{z}$ to be continuous on $\mathbb{C}$, let alone holomorphic.
To see that there is no holomorphic square root, assume that $f(z)$ is one. Then $f(z)^2 = z$ for all $z$, so $$2f'(z)f(z) = 1$$ for all $z$, but since $f(0) = 0$ (the only possible choice of square root of $0$), this is a contradiction.
On the other hand, if $\Omega$ is a simply connected subset of $\mathbb{C}$ not containing $0$, there is a (in fact two) holomorphic square root, given by $$\sqrt{z} = \exp(\tfrac12\log z)$$ where $\log z$ is a holomorphic logarithm on $\Omega$ (which exists bacuse of the assumption, see for example this question. For example you may take $\Omega$ as the complex plane minus the negative real axis and $\log$ as the principal branch of the logarithm.