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The conjugate of sin(z)

  1. Sep 27, 2012 #1
    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?
     
    Last edited: Sep 28, 2012
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  3. Sep 27, 2012 #2

    Mark44

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    Use the fact that eix = cos(x) + isin(x)
     
  4. Sep 28, 2012 #3
    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].
     
    Last edited: Sep 28, 2012
  5. Sep 28, 2012 #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?
     
  6. Sep 28, 2012 #5

    Curious3141

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    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].
     
  7. Sep 28, 2012 #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
     
  8. Sep 28, 2012 #7

    Curious3141

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    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.
     
    Last edited: Sep 28, 2012
  9. Sep 28, 2012 #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 .
     
  10. Sep 29, 2012 #9

    vela

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    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.
     
  11. Sep 29, 2012 #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
     
  12. Sep 29, 2012 #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?
     
  13. Sep 29, 2012 #12
    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]
     
  14. Sep 29, 2012 #13

    Curious3141

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    This is wrong.

    Right, but irrelevant.

    Also wrong (denominator should be 2i).
     
  15. Sep 29, 2012 #14
    [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.
     
    Last edited: Sep 29, 2012
  16. Sep 30, 2012 #15

    ehild

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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
     
  17. Sep 30, 2012 #16

    Curious3141

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    Ha ha, I guessed it had to be a "weird" function like that! :tongue:
     
  18. Sep 30, 2012 #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
     
  19. Sep 30, 2012 #18

    Curious3141

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    Umm... "conjugal functions"? They're loads of fun, just like conjugal visits in prison. :tongue2:
     
  20. Oct 1, 2012 #19
    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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