When is composition of meromorphic functions meromorphic

Assume that $f$ and $g$ are meromorphic in $\Bbb C$, and that the composition $h = f \circ g$ is also meromorphic in $\Bbb C$.

If $f$ is not a rational function then it has an essential singularity at $z= \infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $\lim_{z\to z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.

Therefore $f \circ g$ is meromorphic in $\Bbb C$ if and only if

• $f$ is a rational function, or
• $g$ is holomorphic in $\Bbb C$.

The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.

I say 'special case' because we can think of a meromorphic function as a holomorphic function $\mathbb C \to P^1(\mathbb C)$.

The converse is often false. In fact, when:

• $f$ is entire and there is a value which it reaches infinitely often
• $g$ is meromorphic with at least one pole

then $f \circ g$ is not meromorphic unless $f$ is constant. Indeed, $f \circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f \circ g$ is constant.

Examples: $f = \sin z$, $g = \frac1z$, or $f = e^z$, $g = \frac1z$, ...