A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 9.1, 9.3

How do I do these? (I converted attempts to an answer.)

(Exer 9.1) Prove $f$ has a zero of multiplicity $m$ at $a \implies \frac 1 f$ has a pole of order $m$ at $a$.

(Exer 9.3) Prove $f$ has an essential singularity at $z_0 \implies \frac 1 f$ has an essential singularity at $z_0$.

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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 $\lim_{z \to z_0} |f(z)| = 0$. Apparently (please help), this contradicts (G2) based on Logan M's answer.

QED Exer 9.3

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