Q:

Analytic function $f(z)$ has zero of order $m$ at $z_0$, order of zero of $f'(z_0)$ and $f^2(z_0)$

(This is HW.) Given $f(z)$ analytic on open disk $D$ containing $z_0$; $f(z)$ has a zero order $m$ at $z = z_0$. I need to find the order of zero at $z_0$ of $f'$ and $f^2$ (I assume $f^2$ is the second derivative of $f$, but I honestly don't know.)
This seems overly trivial to me. By plugging in the definition of the order of zeros, I easily obtain that $f'(z_0)$ has order $m-1$ and $f^2(z_0)$ with order $m - 2$ (just a change in index after the definition plugging). Am I missing something obvious?

A:

Your approach seems fine, but I suspect that $f^2(z)=\bigl(f(z)\bigr)^2$. In that case, $f^2$ has a zero of order $2m$ at $z_0$.

