Is the Complex Conjugate of Sin Equal to Sin of the Conjugate?

[Pythagorean]

Homework Statement

does $$(\sin{z})^* = \sin{z^*}$$?

(where z is a complex number)

Homework Equations

$$\sin{z} = \frac{1}{2} (e^{iz} - e^{-iz})$$

The Attempt at a Solution

$$(\sin{z})^* = \frac{1}{2} (e^{iz} - e^{-iz})^*<br /> = \frac{1}{2} (e^{-iz^*} - e^{iz^*}) <br /> = -\frac{1}{2} (e^{iz^*} - e^{-iz^*})<br /> = -sin(z^*)$$

...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.

 

[HallsofIvy]

Here's your error:
$$\sin{z} = \frac{1}{2} (e^{iz} - e^{-iz})$$
is wrong.

$$\sin{z} = \frac{1}{2i} (e^{iz} - e^{-iz})$$


It's the "i" in the denominator that will take care of that sign.

 

[Pythagorean]

ah, right, old habit from the days of real.

Thank you HallsofIvy