# Complex function conjugate

## 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?

## 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?

AB

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.

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

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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

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