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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.

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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.

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