Real part of a function [closed]

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Suppose we have the complex number $c$ and a rationale function $F(c)$. Could I say that $F(mathrmRe(c))=mathrmRe(F(c))$. Is there a proof of this?

Thank you!

linear-algebra algebra-precalculus functions complex-numbers

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edited Jul 20 at 14:11

M. Winter

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asked Jul 20 at 14:09

user9952796

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closed as off-topic by amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel Jul 21 at 1:52

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." – amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
  – user577471
  Jul 20 at 14:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
  – user577471
  Jul 20 at 14:14
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Suppose we have the complex number $c$ and a rationale function $F(c)$. Could I say that $F(mathrmRe(c))=mathrmRe(F(c))$. Is there a proof of this?

Thank you!

linear-algebra algebra-precalculus functions complex-numbers

share|cite|improve this question

edited Jul 20 at 14:11

M. Winter

17.7k62764

asked Jul 20 at 14:09

user9952796

112

closed as off-topic by amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel Jul 21 at 1:52

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." – amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
  – user577471
  Jul 20 at 14:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
  – user577471
  Jul 20 at 14:14
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Suppose we have the complex number $c$ and a rationale function $F(c)$. Could I say that $F(mathrmRe(c))=mathrmRe(F(c))$. Is there a proof of this?

Thank you!

linear-algebra algebra-precalculus functions complex-numbers

share|cite|improve this question

edited Jul 20 at 14:11

M. Winter

17.7k62764

asked Jul 20 at 14:09

user9952796

112

Suppose we have the complex number $c$ and a rationale function $F(c)$. Could I say that $F(mathrmRe(c))=mathrmRe(F(c))$. Is there a proof of this?

Thank you!

linear-algebra algebra-precalculus functions complex-numbers

share|cite|improve this question

edited Jul 20 at 14:11

M. Winter

17.7k62764

asked Jul 20 at 14:09

user9952796

112

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 14:11

M. Winter

17.7k62764

edited Jul 20 at 14:11

M. Winter

17.7k62764

edited Jul 20 at 14:11

M. Winter

17.7k62764

17.7k62764

asked Jul 20 at 14:09

user9952796

112

asked Jul 20 at 14:09

user9952796

112

asked Jul 20 at 14:09

user9952796

112

112

closed as off-topic by amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel Jul 21 at 1:52

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." – amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel

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

closed as off-topic by amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel Jul 21 at 1:52

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." – amWhy, Strants, jgon, Taroccoesbrocco, Parcly Taxel

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
  – user577471
  Jul 20 at 14:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
  – user577471
  Jul 20 at 14:14
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
  – user577471
  Jul 20 at 14:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
  – user577471
  Jul 20 at 14:14
  
  

  
  
  
  

Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
– user577471
Jul 20 at 14:10

Take $F(c)=ic$. Then $F(Re(c))=iRe(c)$, while $Re(F(c))=Re(ic)=-Im(c)$.
– user577471
Jul 20 at 14:10

Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
– user577471
Jul 20 at 14:14

Even with real coefficients $F(c)=1/c$ gives $F(Re(c))=1/Re(c)$, while $Re(F(c))=Re(1/c)=Re(c)/|c|$
– user577471
Jul 20 at 14:14

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$DeclareMathOperatorReReRe(F(z))=0not=i=F(Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$Re(F(i))=-1not=0=F(Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)inBbb R$ as well. Therefore

$$F(Re(x))=underbraceF(x)_in ,Bbb R=Re(F(x)).$$

share|cite|improve this answer

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$DeclareMathOperatorReReRe(F(z))=0not=i=F(Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$Re(F(i))=-1not=0=F(Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)inBbb R$ as well. Therefore

$$F(Re(x))=underbraceF(x)_in ,Bbb R=Re(F(x)).$$

share|cite|improve this answer

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

add a comment | 

up vote
1
down vote



accepted

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$DeclareMathOperatorReReRe(F(z))=0not=i=F(Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$Re(F(i))=-1not=0=F(Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)inBbb R$ as well. Therefore

$$F(Re(x))=underbraceF(x)_in ,Bbb R=Re(F(x)).$$

share|cite|improve this answer

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$DeclareMathOperatorReReRe(F(z))=0not=i=F(Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$Re(F(i))=-1not=0=F(Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)inBbb R$ as well. Therefore

$$F(Re(x))=underbraceF(x)_in ,Bbb R=Re(F(x)).$$

share|cite|improve this answer

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$DeclareMathOperatorReReRe(F(z))=0not=i=F(Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$Re(F(i))=-1not=0=F(Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)inBbb R$ as well. Therefore

$$F(Re(x))=underbraceF(x)_in ,Bbb R=Re(F(x)).$$

share|cite|improve this answer

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 14:31

edited Jul 20 at 14:31

edited Jul 20 at 14:31

answered Jul 20 at 14:25

M. Winter

17.7k62764

answered Jul 20 at 14:25

M. Winter

17.7k62764

answered Jul 20 at 14:25

M. Winter

17.7k62764

17.7k62764

add a comment | 

add a comment |