$f$ zero (essential singularity) $\implies \frac 1 f$ pole (essential singularity)

Pf of Exer 9.1: By Classification of Zeroes Thm 8.14, $ f \equiv 0$ or $f=(z-z_0)^mg(z)$. If the former, then the conclusion holds true vacuously. If the latter, then $$\frac 1 f = \frac 1{(z-z_0)^mg(z)}=\frac {\frac1{g(z)}}{(z-z_0)^m}$$

$\therefore, \frac1f$ satisfies the conditions of Cor 9.6 of Prop 9.5. I omit specific details of the conditions that $g$ has and that $\frac1g$ satisfies. QED Exer 9.1

Pf of Exer 9.3

We are given $f$

• (G1) has a singularity: $f$ is holomorphic on $0<|z-z_0|<R, R>0$, but not holomorphic at $z=z_0$

• (G2) which is not removable: $\nexists g$ holomorphic in $|z-z_0|<R$ s.t. $f=g$ on $0<|z-z_0|<R$

• (G3) and not a pole: $\lim_{z \to z_0} |f(z)| \ne \infty$

• (G4) and f is not zero: $f \not\equiv 0 \ \forall z \in |z-z_0|<R$

We must show $\frac 1f$

• (S1) has a singularity: $\frac 1f$ is holomorphic on $0<|z-z_0|<R_1, R_1>0$, but not holomorphic at $z=z_0$.

• (S2) which is not removable: $\nexists h$ holomorphic in $|z-z_0|<R_1$ s.t. $\frac 1 f=h$ on $0<|z-z_0|<R_1$

• (S3) and not a pole: $\lim_{z \to z_0} |\frac1{f(z)}| \ne \infty$

Now:

• (S1) follows from (G1) and (G4).
• (S2) follows from (G2).
• For (S3), suppose on the contrary that $\lim_{z \to z_0} |\frac1{f(z)}| = \infty$. Then $\lim_{z \to z_0} |f(z)| = 0$. Ostensibly (please help), this contradicts (G2) based on Logan M's answer.

QED Exer 9.3