STRACT: Understanding the Complex Conjugate of a Function

[leoneri]

Homework Statement

I have a complex function

w\left(z\right)=e^{sin\left(z\right)}

What is the conjugate?

2. The attempt at a solution

The conjugate is

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

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?

 

[Altabeh]

Homework Statement

I have a complex function

w\left(z\right)=e^{sin\left(z\right)}

What is the conjugate?

2. The attempt at a solution

The conjugate is

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

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

 

[leoneri]

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 w\left(z\right)=w\left(x+iy}\right), then the conjugate is always w\left(z^{*}\right)=w\left(x-iy\right). I would be more than happy if someone can show me or point me to a proof that the above definition is always true.

 

[Rjee]

But I think you need
W*(z)
and not
W(z*)

Ask your advisor B. ...L.P. !

 

[Altabeh]

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 w\left(z\right)=w\left(x+iy}\right), then the conjugate is always w\left(z^{*}\right)=w\left(x-iy\right). 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 e^{i\theta}. In this case, using the Euler relation one can see

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

so

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

Here you can understand that the result of the product of the function e^{i\theta} 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 e^{i\theta} and e^{-i\theta}. Thus a change in the sign of i makes the function be transferred into its complex conjugate phase.

AB