What is the Conjugate of sin(z)?

[J.L.A.N.]

The problem is to show $sin\overline{z} = \overline{sinz}$. What I need is help to get going.We know that $sinz = \frac{e^{iz}-e^{-iz}}{2i}$I can't see the first step in this. What I've tried to do is expressing $sin\overline{z}$ and $\overline{sinz}$ in terms of the above equation, but I don't know how to write the conjugate of $\frac{e^{iz}-e^{-iz}}{2i}$. Of course I know the conjugate of a regular complex number, $\overline{z} = x - iy$. How do these relate?

 

[Mark44]

Use the fact that e^ix = cos(x) + isin(x)

 

[J.L.A.N.]

Indeed, that's what I did. But it doesn't take me anywhere: What I get is $\overline{sinz} = \overline{\frac{cosz+isinz-cos(-z)-isin(-z)}{2i}}$. And I don't see how I could turn that into the form $\overline{z} = x-iy$. Let alone $\frac{cos\overline{z}+isin\overline{z}-cos(-\overline{z})-isin(-\overline{z})}{2i} (= sin\overline{z})$.

 

[Mentallic]

Can you simplify cos(-z)? What about sin(-z)?

Now, you know that the conjugate of a complex number a+ib is a-ib, so what you need to do is to make sure you always convert it into the form a+ib, remembering that a and b can any complicated expression, as long as you have the real numbers on one side (a) and then the imaginary numbers (b) all multiplied by i.

So for example, if I had 1+ix+e^x-i.cos(x) then to find the conjugate I'd first group all real terms and then all imaginary terms like so:

(1+e^x)+i(x-cos(x))

Which then the conjugate is

(1+e^x)-i(x-cos(x))

So for your problem, you also have the issue that i is in the denominator. How can we get rid of that?

 

[Curious3141]

Indeed, that's what I did. But it doesn't take me anywhere: What I get is $\overline{sinz} = \overline{\frac{cosz+isinz-cos(-z)-isin(-z)}{2i}}$. And I don't see how I could turn that into the form $\overline{z} = x-iy$. Let alone $\frac{cos\overline{z}+isin\overline{z}-cos(-\overline{z})-isin(-\overline{z})}{2i} (= sin\overline{z})$.

This line of thinking is not going to get you anywhere, because after simplification, you're going to end up with the trivial identity $\sin z = \sin z$.

Better to start with $z = x + yi$, where x and y are real, then use the angle sum identity for sine to work out $\overline{\sin z}$ and $\sin{\overline{z}}$ and prove they're equal. You'll need to use hyperbolic trig functions to simplify the circular trig ratios for the imaginary part $yi$.

 

[ehild]

You do not need to use z=x+iy. Start with the conjugate of sin(z). You get the conjugate by changing i to -i and z to $\bar {z}$.

ehild

 

[Curious3141]

You do not need to use z=x+iy. Start with the conjugate of sin(z). You get the conjugate by changing i to -i and z to $\bar {z}$.

ehild

How exactly would one simplify sin(z) here, then get its conjugate?

EDIT: I guess you mean he would apply this step to: $\sin z = \frac{e^{iz} - e^{-iz}}{2i}$?

Then my question is: can you apply such an operation without a general proof?

It would be sufficient to prove that $\overline{e^{iz}} = e^{-i\overline{z}}$. But as far as I know, this is not a result that can be assumed without proof.

 

[ehild]

Well, when I studied complex numbers it was shown to us that we get the conjugate of anything by changing i to -i and the variables to their conjugate. Anyway, the conjugate of a sum is the sum of conjugate, and it is the same with a product. e^iz was defined with its Taylor expansion, which contains products and a sum.

But the conjugate of e^iz is easy to derive.

$$e^{iz}=e^{i(x+iy)}=e^{ix-y}=e^{-y}(cosx+isinx)$$

$$\overline{e^{iz}}=\overline{e^{-y}(cosx+isinx)}=e^{-y}(cosx-isinx)=e^{-y}(cos(-x)+isin(-x))=e^{-y-ix}=e^{-i(x-iy)}=e^{-i\bar{z}}$$

ehild .

 

[vela]

Then my question is: can you apply such an operation without a general proof?

Depends on the class, I suppose. My first thought was the same as ehild's, but only the OP can tell us if that would be an appropriate proof.

 

[ehild]

I think the simple proof I have shown is taught when the function e^iz is defined in classes.


ehild

 

[Curious3141]

OK, thanks, I guess it's up to the poster to tell us if he can actually use that without proof.

By the way, when does $\overline{f(z)} = f({\overline{z}})$? It's certainly true for all polynomial functions. By extension, I guess any function that has a Taylor series also satisfies this criterion. Can you think of any counterexamples?

 

[help1please]

The problem is to show $sin\overline{z} = \overline{sinz}$. What I need is help to get going.


We know that $sinz = \frac{e^{iz}-e^{-iz}}{2i}$


I can't see the first step in this. What I've tried to do is expressing $sin\overline{z}$ and $\overline{sinz}$ in terms of the above equation, but I don't know how to write the conjugate of $\frac{e^{iz}-e^{-iz}}{2i}$. Of course I know the conjugate of a regular complex number, $\overline{z} = x - iy$. How do these relate?

You want to know how to find the conjugate of $sinz = \frac{e^{iz}-e^{-iz}}{2i}$

I'd do it as

$$sinz* = \frac{e^{iz}+e^{-iz}}{2i}$$

all you do is change the sign. It is similar to

$$cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$$

$$sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2}$$

 

[Curious3141]

You want to know how to find the conjugate of $sinz = \frac{e^{iz}-e^{-iz}}{2i}$

I'd do it as

$$sinz* = \frac{e^{iz}+e^{-iz}}{2i}$$

all you do is change the sign.

This is wrong.

It is similar to

$$cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$$

Right, but irrelevant.

$$sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2}$$

Also wrong (denominator should be 2i).

 

[jackmell]

but I don't know how to write the conjugate of $\frac{e^{iz}-e^{-iz}}{2i}$.

$$\overline{\left(\frac{z}{w}\right)} =\frac{\overline{z}}{\overline{w}}$$

and:

$$\overline{z-w}=\overline{z}-\overline{w}$$

There you go then:

$$\overline{\left(\frac{e^{iz}-e^{-iz}}{2i}\right)}= \frac{\overline{e^{iz}-e^{-iz}}}{\overline{2i}}= \frac{\overline{\left(e^{iz}\right)}-\overline{\left(e^{-iz}\right)}}{-2i}$$

and it's not hard to musscle-through the conjugate of the exponents:

$$\overline{\left(e^{iz}\right)}=\overline{e^{-b}\left(\cos(a)-i\sin(a)\right)}=e^{\left(\overline{iz}\right)}=e^{-i\overline{z}}$$

You can finish it.

Edit: Sorry, just saw others did similar and can't delete it.

 

[ehild]

By the way, when does $\overline{f(z)} = f({\overline{z}})$? It's certainly true for all polynomial functions. By extension, I guess any function that has a Taylor series also satisfies this criterion. Can you think of any counterexamples?

What about f(z)=1 if Im(z)≥0 and f(z)=0 if Im(z)<0? The conjugate of f(z) is itself, but f*(u+iv)=f(u+iv)≠f(u-iv).

ehild

 

[Curious3141]

What about f(z)=1 if Im(z)≥0 and f(z)=0 if Im(z)<0? The conjugate of f(z) is itself, but f*(u+iv)=f(u+iv)≠f(u-iv).

ehild

Ha ha, I guessed it had to be a "weird" function like that! :tongue:

 

[ehild]

Well, my Math knowledge is very rusty. What can be the name of those functions for which f*(z)=f(z*)?

ehild

 

[Curious3141]

Well, my Math knowledge is very rusty. What can be the name of those functions for which f*(z)=f(z*)?

ehild

Umm... "conjugal functions"? They're loads of fun, just like conjugal visits in prison. :tongue2:

 

[help1please]

This is wrong.



Right, but irrelevant.



Also wrong (denominator should be 2i).

I apologize, there should have been an imaginary number in the denominator.

However I don't see how it is irrelevant. If they want to conjugate the equation they gave, we are talking about a simple sign change. Why would that be wrong?