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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.