Complex function conjugate

  • Thread starter leoneri
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  • #1
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Homework Statement



I have a complex function

[tex]w\left(z\right)=e^{sin\left(z\right)}[/tex]

What is the conjugate?

2. The attempt at a solution

The conjugate is

[tex]w\left(z^{*}\right)=e^{sin\left(z^{*}\right)}[/tex]
[tex]w\left(x-iy\right)=e^{sin\left(x-iy\right)}[/tex]

My question is, is my answer correct? If I wrong, could you tell me what is the right answer and how to reach this correct answer?
 

Answers and Replies

  • #2
660
0

Homework Statement



I have a complex function

[tex]w\left(z\right)=e^{sin\left(z\right)}[/tex]

What is the conjugate?

2. The attempt at a solution

The conjugate is

[tex]w\left(z^{*}\right)=e^{sin\left(z^{*}\right)}[/tex]
[tex]w\left(x-iy\right)=e^{sin\left(x-iy\right)}[/tex]

My question is, is my answer correct? If I wrong, could you tell me what is the right answer and how to reach this correct answer?

Your answer is flawless!

AB
 
  • #3
19
0
Thanks. I was sure about that, but one of my friend said I was wrong. So, is there a way to proof it? Because I see it as a definition that whenever there is a complex function [tex]w\left(z\right)=w\left(x+iy}\right)[/tex], then the conjugate is always [tex]w\left(z^{*}\right)=w\left(x-iy\right)[/tex]. I would be more than happy if someone can show me or point me to a proof that the above definition is always true.
 
  • #4
1
0
But I think you need
W*(z)
and not
W(z*)

Ask your advisor B. .....L.P. !!
 
Last edited:
  • #5
660
0
Thanks. I was sure about that, but one of my friend said I was wrong. So, is there a way to proof it? Because I see it as a definition that whenever there is a complex function [tex]w\left(z\right)=w\left(x+iy}\right)[/tex], then the conjugate is always [tex]w\left(z^{*}\right)=w\left(x-iy\right)[/tex]. I would be more than happy if someone can show me or point me to a proof that the above definition is always true.

Since that is a definition, we can't do so much to give a real proof. But you can imagine the case involving [tex]e^{i\theta}[/tex]. In this case, using the Euler relation one can see

[tex]e^{i\theta}=\cos(\theta)+i\sin(\theta)[/tex],
[tex]e^{i(-\theta)}=\cos(\theta)-i\sin(\theta)[/tex] (by the symmetry of cos and anti-symmetry of sin wrt the change of the sign of [tex]\theta[/tex]),

so

[tex]e^{i\theta}e^{i(-\theta)}=[\cos(\theta)+i\sin(\theta)][\cos(\theta)-i\sin(\theta)] = 1.[/tex].

Here you can understand that the result of the product of the function [tex]e^{i\theta}[/tex] and its complex conjugate leads to the correct answer 1 just by looking at the first part of the equality. The second part confirms the complex conjugation of [tex]e^{i\theta}[/tex] and [tex]e^{-i\theta}[/tex]. Thus a change in the sign of [tex]i[/tex] makes the function be transferred into its complex conjugate phase.

AB
 
Last edited:

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