Complex $f$ is constant if $Re(f)=text{constant}$ or $Im(f)=text{constant}$ [on hold]

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Claim: Let $f$ be complex differentiable. If $Re(f)=text{constant}$ or $Im(f)=text{constant}$, then $f$ is constant.

How do I prove that? I can only see how to do it if $f'$ would be continuous but this isn't the case here. So how do I go about it?

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

put on hold as off-topic by Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer

If this question can be reworded to fit the rules in the help center, please edit the question.

1

With a simple search in this site you can find one dozen answers for your question.
– Nosrati
2 days ago

I did some searching but in all such questions $f$ was analytic/holomorphic.
– xotix
2 days ago

1

@xotic of course, because if $f$ is complex-differentiable then it is holomorphic, what is the same to say that it is analytic
– Masacroso
2 days ago

@Masacroso my script clearly states that $f$ is said to be (complex) analytic, if $f$ is differentiable and $f'$ is continous.
– xotix
2 days ago

1

@xotic as I said: if $f$ is complex-differentiable then indeed $f$ is holomorphic and $f'$ is continuous
– Masacroso
2 days ago

|
show 1 more comment

up vote
-3
down vote

favorite

Claim: Let $f$ be complex differentiable. If $Re(f)=text{constant}$ or $Im(f)=text{constant}$, then $f$ is constant.

How do I prove that? I can only see how to do it if $f'$ would be continuous but this isn't the case here. So how do I go about it?

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

put on hold as off-topic by Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer

If this question can be reworded to fit the rules in the help center, please edit the question.

1

With a simple search in this site you can find one dozen answers for your question.
– Nosrati
2 days ago

I did some searching but in all such questions $f$ was analytic/holomorphic.
– xotix
2 days ago

1

@xotic of course, because if $f$ is complex-differentiable then it is holomorphic, what is the same to say that it is analytic
– Masacroso
2 days ago

@Masacroso my script clearly states that $f$ is said to be (complex) analytic, if $f$ is differentiable and $f'$ is continous.
– xotix
2 days ago

1

@xotic as I said: if $f$ is complex-differentiable then indeed $f$ is holomorphic and $f'$ is continuous
– Masacroso
2 days ago

|
show 1 more comment

up vote
-3
down vote

favorite

Claim: Let $f$ be complex differentiable. If $Re(f)=text{constant}$ or $Im(f)=text{constant}$, then $f$ is constant.

How do I prove that? I can only see how to do it if $f'$ would be continuous but this isn't the case here. So how do I go about it?

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

Claim: Let $f$ be complex differentiable. If $Re(f)=text{constant}$ or $Im(f)=text{constant}$, then $f$ is constant.

How do I prove that? I can only see how to do it if $f'$ would be continuous but this isn't the case here. So how do I go about it?

complex-analysis

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

edited 2 days ago

Tianlalu

2,210631

edited 2 days ago

Tianlalu

2,210631

edited 2 days ago

Tianlalu

2,210631

asked 2 days ago

xotix

31829

asked 2 days ago

xotix

31829

asked 2 days ago

xotix

31829

put on hold as off-topic by Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, T. Bongers, rtybase, ArsenBerk, Chris Custer

If this question can be reworded to fit the rules in the help center, please edit the question.

1

With a simple search in this site you can find one dozen answers for your question.
– Nosrati
2 days ago

I did some searching but in all such questions $f$ was analytic/holomorphic.
– xotix
2 days ago

1

@xotic of course, because if $f$ is complex-differentiable then it is holomorphic, what is the same to say that it is analytic
– Masacroso
2 days ago

@Masacroso my script clearly states that $f$ is said to be (complex) analytic, if $f$ is differentiable and $f'$ is continous.
– xotix
2 days ago

1

@xotic as I said: if $f$ is complex-differentiable then indeed $f$ is holomorphic and $f'$ is continuous
– Masacroso
2 days ago

|
show 1 more comment

1

With a simple search in this site you can find one dozen answers for your question.
– Nosrati
2 days ago

I did some searching but in all such questions $f$ was analytic/holomorphic.
– xotix
2 days ago

1

@xotic of course, because if $f$ is complex-differentiable then it is holomorphic, what is the same to say that it is analytic
– Masacroso
2 days ago

@Masacroso my script clearly states that $f$ is said to be (complex) analytic, if $f$ is differentiable and $f'$ is continous.
– xotix
2 days ago

1

@xotic as I said: if $f$ is complex-differentiable then indeed $f$ is holomorphic and $f'$ is continuous
– Masacroso
2 days ago

With a simple search in this site you can find one dozen answers for your question.
– Nosrati
2 days ago

I did some searching but in all such questions $f$ was analytic/holomorphic.
– xotix
2 days ago

@xotic of course, because if $f$ is complex-differentiable then it is holomorphic, what is the same to say that it is analytic
– Masacroso
2 days ago

@Masacroso my script clearly states that $f$ is said to be (complex) analytic, if $f$ is differentiable and $f'$ is continous.
– xotix
2 days ago

@xotic as I said: if $f$ is complex-differentiable then indeed $f$ is holomorphic and $f'$ is continuous
– Masacroso
2 days ago

|
show 1 more comment

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