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The conjugate of sin(z)

  • Thread starter J.L.A.N.
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  • #1
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The problem is to show [itex]sin\overline{z} = \overline{sinz}[/itex]. What I need is help to get going.


We know that [itex]sinz = \frac{e^{iz}-e^{-iz}}{2i}[/itex]


I can't see the first step in this. What I've tried to do is expressing [itex]sin\overline{z} [/itex] and [itex] \overline{sinz} [/itex] in terms of the above equation, but I don't know how to write the conjugate of [itex]\frac{e^{iz}-e^{-iz}}{2i}[/itex]. Of course I know the conjugate of a regular complex number, [itex]\overline{z} = x - iy[/itex]. How do these relate?
 
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  • #2
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Use the fact that eix = cos(x) + isin(x)
 
  • #3
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Indeed, that's what I did. But it doesn't take me anywhere: What I get is [itex] \overline{sinz} = \overline{\frac{cosz+isinz-cos(-z)-isin(-z)}{2i}} [/itex]. And I don't see how I could turn that into the form [itex] \overline{z} = x-iy [/itex]. Let alone [itex] \frac{cos\overline{z}+isin\overline{z}-cos(-\overline{z})-isin(-\overline{z})}{2i} (= sin\overline{z}) [/itex].
 
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  • #4
Mentallic
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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+ex-i.cos(x) then to find the conjugate I'd first group all real terms and then all imaginary terms like so:

(1+ex)+i(x-cos(x))

Which then the conjugate is

(1+ex)-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?
 
  • #5
Curious3141
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Indeed, that's what I did. But it doesn't take me anywhere: What I get is [itex] \overline{sinz} = \overline{\frac{cosz+isinz-cos(-z)-isin(-z)}{2i}} [/itex]. And I don't see how I could turn that into the form [itex] \overline{z} = x-iy [/itex]. Let alone [itex] \frac{cos\overline{z}+isin\overline{z}-cos(-\overline{z})-isin(-\overline{z})}{2i} (= sin\overline{z}) [/itex].
This line of thinking is not going to get you anywhere, because after simplification, you're going to end up with the trivial identity [itex]\sin z = \sin z[/itex].

Better to start with [itex]z = x + yi[/itex], where x and y are real, then use the angle sum identity for sine to work out [itex]\overline{\sin z}[/itex] and [itex]\sin{\overline{z}}[/itex] and prove they're equal. You'll need to use hyperbolic trig functions to simplify the circular trig ratios for the imaginary part [itex]yi[/itex].
 
  • #6
ehild
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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 [itex]\bar {z}[/itex].

ehild
 
  • #7
Curious3141
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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 [itex]\bar {z}[/itex].

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

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

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

It would be sufficient to prove that [itex]\overline{e^{iz}} = e^{-i\overline{z}}[/itex]. But as far as I know, this is not a result that can be assumed without proof.
 
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  • #8
ehild
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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. eiz was defined with its Taylor expansion, which contains products and a sum.

But the conjugate of eiz is easy to derive.

[tex]e^{iz}=e^{i(x+iy)}=e^{ix-y}=e^{-y}(cosx+isinx)[/tex]

[tex]\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}}[/tex]

ehild .
 
  • #9
vela
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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.
 
  • #10
ehild
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I think the simple proof I have shown is taught when the function eiz is defined in classes.


ehild
 
  • #11
Curious3141
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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 [itex]\overline{f(z)} = f({\overline{z}})[/itex]? 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?
 
  • #12
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The problem is to show [itex]sin\overline{z} = \overline{sinz}[/itex]. What I need is help to get going.


We know that [itex]sinz = \frac{e^{iz}-e^{-iz}}{2i}[/itex]


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

I'd do it as

[tex]sinz* = \frac{e^{iz}+e^{-iz}}{2i}[/tex]

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

[tex]cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex]

[tex]sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2}[/tex]
 
  • #13
Curious3141
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You want to know how to find the conjugate of [itex]sinz = \frac{e^{iz}-e^{-iz}}{2i}[/itex]

I'd do it as

[tex]sinz* = \frac{e^{iz}+e^{-iz}}{2i}[/tex]

all you do is change the sign.
This is wrong.

It is similar to

[tex]cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex]
Right, but irrelevant.

[tex]sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2}[/tex]
Also wrong (denominator should be 2i).
 
  • #14
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but I don't know how to write the conjugate of [itex]\frac{e^{iz}-e^{-iz}}{2i}[/itex].
[tex]\overline{\left(\frac{z}{w}\right)} =\frac{\overline{z}}{\overline{w}}[/tex]

and:

[tex]\overline{z-w}=\overline{z}-\overline{w}[/tex]

There you go then:

[tex]\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}[/tex]

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

[tex]\overline{\left(e^{iz}\right)}=\overline{e^{-b}\left(\cos(a)-i\sin(a)\right)}=e^{\left(\overline{iz}\right)}=e^{-i\overline{z}}[/tex]

You can finish it.

Edit: Sorry, just saw others did similar and can't delete it.
 
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  • #15
ehild
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By the way, when does [itex]\overline{f(z)} = f({\overline{z}})[/itex]? 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
 
  • #16
Curious3141
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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:
 
  • #17
ehild
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Well, my Math knowledge is very rusty. What can be the name of those functions for which f*(z)=f(z*)?

ehild
 
  • #18
Curious3141
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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:
 
  • #19
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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?
 

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