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$f$ zero (essential singularity) $implies frac 1 f$ pole (essential singularity)
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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 $z_0$.
complex-analysis limits derivatives proof-verification residue-calculus
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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 $z_0$.
complex-analysis limits derivatives proof-verification residue-calculus
asked Aug 6 at 16:05
BCLC
6,77021973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-20 09:57:41Z">in 20 hours.
This question has not received enough attention.
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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 $z_0$.
complex-analysis limits derivatives proof-verification residue-calculus
asked Aug 6 at 16:05
BCLC
6,77021973
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 $z_0$.
complex-analysis limits derivatives proof-verification residue-calculus
asked Aug 6 at 16:05
BCLC
6,77021973
edited Aug 15 at 10:59
asked Aug 6 at 16:05
BCLC
6,77021973
asked Aug 6 at 16:05
BCLC
6,77021973
asked Aug 6 at 16:05
BCLC
6,77021973
6,77021973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-20 09:57:41Z">in 20 hours.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-20 09:57:41Z">in 20 hours.
This question has not received enough attention.
add a comment |Â
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1 Answer
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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 $$frac 1 f = frac 1(z-z_0)^mg(z)=frac frac1g(z)(z-z_0)^m$$
$therefore, frac1f$ 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: $f notequiv 0 forall z in |z-z_0|<R$
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: $lim_z to z_0 |frac1f(z)| ne infty$
Now:
- (S1) follows from (G1) and (G4).
- (S2) follows from (G2).
- For (S3), suppose on the contrary that $lim_z to z_0 |frac1f(z)| = infty$. 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
answered Aug 15 at 7:28
BCLC
6,77021973
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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 $$frac 1 f = frac 1(z-z_0)^mg(z)=frac frac1g(z)(z-z_0)^m$$
$therefore, frac1f$ 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: $f notequiv 0 forall z in |z-z_0|<R$
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: $lim_z to z_0 |frac1f(z)| ne infty$
Now:
- (S1) follows from (G1) and (G4).
- (S2) follows from (G2).
- For (S3), suppose on the contrary that $lim_z to z_0 |frac1f(z)| = infty$. 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
answered Aug 15 at 7:28
BCLC
6,77021973
add a comment |Â
up vote
1
down vote
accepted
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 $$frac 1 f = frac 1(z-z_0)^mg(z)=frac frac1g(z)(z-z_0)^m$$
$therefore, frac1f$ 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: $f notequiv 0 forall z in |z-z_0|<R$
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: $lim_z to z_0 |frac1f(z)| ne infty$
Now:
- (S1) follows from (G1) and (G4).
- (S2) follows from (G2).
- For (S3), suppose on the contrary that $lim_z to z_0 |frac1f(z)| = infty$. 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
answered Aug 15 at 7:28
BCLC
6,77021973
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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 $$frac 1 f = frac 1(z-z_0)^mg(z)=frac frac1g(z)(z-z_0)^m$$
$therefore, frac1f$ 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: $f notequiv 0 forall z in |z-z_0|<R$
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: $lim_z to z_0 |frac1f(z)| ne infty$
Now:
- (S1) follows from (G1) and (G4).
- (S2) follows from (G2).
- For (S3), suppose on the contrary that $lim_z to z_0 |frac1f(z)| = infty$. 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
answered Aug 15 at 7:28
BCLC
6,77021973
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 $$frac 1 f = frac 1(z-z_0)^mg(z)=frac frac1g(z)(z-z_0)^m$$
$therefore, frac1f$ 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: $f notequiv 0 forall z in |z-z_0|<R$
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: $lim_z to z_0 |frac1f(z)| ne infty$
Now:
- (S1) follows from (G1) and (G4).
- (S2) follows from (G2).
- For (S3), suppose on the contrary that $lim_z to z_0 |frac1f(z)| = infty$. 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
answered Aug 15 at 7:28
BCLC
6,77021973
answered Aug 15 at 7:28
BCLC
6,77021973
answered Aug 15 at 7:28
BCLC
6,77021973
answered Aug 15 at 7:28
BCLC
6,77021973
6,77021973
add a comment |Â
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