# Complex Conjugate of Sin

1. Jan 29, 2008

### Pythagorean

1. The problem statement, all variables and given/known data

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

(where z is a complex number)

2. Relevant equations

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

3. The attempt at a solution

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

...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.

2. Jan 29, 2008

### HallsofIvy

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

3. Jan 29, 2008

### Pythagorean

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

Thank you HallsofIvy