# If $f$ has a zero of order $k$ at $a$ and $g$ has a pole of order $m$ at $z=0$, what does $g(f(z))$ have at $z=a$?

$f,g$ be meromorphic on $\mathbb{C}$, $f$ has a zero of order $k$ at $a$ and $g$ has a pole of order $m$ at $z=0$, then $g(f(z))$ has

1. a zero of order $km$ at $z=a$

2. a pole of order $km$ at $z=a$

3. a zero of order $|k-m|$ at $z=a$

4. a pole of order $|k-m|$ at $z=a$

Could any one please tell me, I am not able to get the answer. Thanks for helping.

• Take the simplest functions with the required properties. Look at what $g\circ f$ is for those. – Daniel Fischer Nov 25 '14 at 18:40

## 2 Answers

Write $f(z)=(z-a)^k\bar f(z)$ and $g(z)=\bar g(z)/z^m$ where $\bar f$ does not vanish at $z=a$ and $\bar g$ is holomorphic and does not vanish at $z=0$.

Then,

$$g(f(z))=\frac1{z^m}\bar g((z-a)^k\bar f(z))$$

which has a pole of order at least $m$ at $z=0$.

It's obvious that $g\circ f$ is not defined at $z=a$, so it has a pole. Since this is true also when $k=m$, possibility number $4$ turns out to be false, and so the answer must be number $2$. Now that you know what the answer is, prove it.