Complex Conjugate of Sin

  • #1
Pythagorean
Gold Member
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276

Homework Statement



does [tex](\sin{z})^* = \sin{z^*}[/tex]?

(where z is a complex number)

Homework Equations



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

The Attempt at a Solution



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

...but my teacher told us ahead of time that they should be equivalent. I'm not seeing my mistake. Thank you for your time and energy.

I will also note that the same problem for cos only worked out to be equivalent because of the commutative rule. This doesn't work for sin, since it's exponential terms differ in their signs.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Here's your error:
[tex]\sin{z} = \frac{1}{2} (e^{iz} - e^{-iz})[/tex]
is wrong.

[tex]\sin{z} = \frac{1}{2i} (e^{iz} - e^{-iz})[/tex]


It's the "i" in the denominator that will take care of that sign.
 
  • #3
Pythagorean
Gold Member
4,291
276
ah, right, old habit from the days of real.

Thank you HallsofIvy
 

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