Complex Trig Identity

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  • #1
kreil
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Homework Statement


Show [tex]|sin z|^2 = \frac{1}{2}[cosh(2y)-cos(2x)][/tex]


Homework Equations


[tex]cosh2y = cosh^2y+sinh^2y[/tex]
[tex]cos2x = cos^2x-sin^2x[/tex]


The Attempt at a Solution



Here is what I have so far

[tex]|sinz|^2=|sin(x+iy)|^2=|sin(x)cosh(y)+icos(x)sinh(y)|^2[/tex]

[tex]=sin^2(x)cosh^2(y)+cos^2(x)sinh^2(y)[/tex]

[tex]=sin^2(x)cosh^2(y)+cos^2(x)sinh^2(y)-sin^2(x)sinh^2(y)+sin^2(x)sinh^2(y)[/tex]

[tex]=sin^2(x)[cosh^2(y)+sinh^2(y)]+sinh^2(y)[cos^2(x)-sin^2(x)][/tex]

[tex]=sin^2(x)cosh(2y)+sinh^2(y)cos(2x)[/tex]

how should i proceed?
 

Answers and Replies

  • #2
vela
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I'd try using the identity

[tex]\cos^2 x = \frac{1-\cos(2x)}{2}[/tex]

I assume there's a similar one for [itex]\sinh^2 y[/itex].
 
  • #3
Dick
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This also works out very directly if you use sin(z)=(exp(iz)-exp(-iz))/(2i), |sin(z)|^2=sin(z)*conjugate(sin(z)) and z=x+iy. If you work directly with the exponentials, you don't need any trig identities.
 
  • #4
kreil
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This also works out very directly if you use sin(z)=(exp(iz)-exp(-iz))/(2i), |sin(z)|^2=sin(z)*conjugate(sin(z)) and z=x+iy. If you work directly with the exponentials, you don't need any trig identities.

Doing it this way I get the following:

[tex]|sinz|^2=sin(z) sin(z)^*= \frac{1}{4} (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})[/tex]

[tex]= \frac{1}{4}(2-e^{2i(x+iy)}-e^{-2i(x+iy)})=\frac{1}{4}(2-e^{i2x}e^{i2iy}-e^{-i2x}e^{-i2iy})[/tex]

Converting to trig fcns,

[tex]\frac{1}{4}[2-(cos2x+isin2x)(cos2iy+isin2iy)-(cos2x-isin2x)(cos2iy-isin2iy)[/tex]

Now I know [itex]sinh(x)=-isin(ix)[/itex], [itex]cosh(x)=cos(ix)[/itex], but am i on the right track? i dont want to expand this out if i've already done something incorrect
 
  • #5
Dick
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Doing it this way I get the following:

[tex]|sinz|^2=sin(z) sin(z)^*= \frac{1}{4} (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})[/tex]

[tex]= \frac{1}{4}(2-e^{2i(x+iy)}-e^{-2i(x+iy)})=\frac{1}{4}(2-e^{i2x}e^{i2iy}-e^{-i2x}e^{-i2iy})[/tex]

Converting to trig fcns,

[tex]\frac{1}{4}[2-(cos2x+isin2x)(cos2iy+isin2iy)-(cos2x-isin2x)(cos2iy-isin2iy)[/tex]

Now I know [itex]sinh(x)=-isin(ix)[/itex], [itex]cosh(x)=cos(ix)[/itex], but am i on the right track? i dont want to expand this out if i've already done something incorrect

Good hunch. Yes, you have done something wrong already. You forgot to change the z to z* in the second factor.
 
  • #6
kreil
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ok ill try that ty
 
Last edited:
  • #7
Dick
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ok ill try that

No, you didn't change z to z*. You wrote exp(-iz)-exp(iz). That should be exp(-iz*)-exp(iz*)=exp(-ix-y)-exp(ix+y). What's the first factor in terms of x and y?
 
  • #8
kreil
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I see now, it was a silly mistake.

[tex]=-\frac{1}{4} \left( e^{i(z+z^*)}-e^{i(z-z^*)}-e^{-i(z-z^*)}+e^{-i(z+z^*)} \right)[/tex]

and since z+z*=2x, z-z*=2iy,

[tex]=-\frac{1}{4} \left( e^{2ix}-e^{-2y}-e^{2y}+e^{-2ix} \right) [/tex]

[tex]=\frac{1}{2} \left( \frac{e^{2y}+e^{-2y}}{2}-\frac{e^{2ix}+e^{-2ix}}{2} \right) [/tex]

[tex]= \frac{1}{2} \left( cosh(2y) - cos(2x) \right) [/tex]

I'm a bit rusty from taking a semester off, thanks for your help!
 

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