# Thread: Trigonometric Function (Complex)

1. ## Trigonometric Function (Complex)

I have 2 problems that I cannot finish/ not sure.

Solve for all values of z such that,
1) $sin(z) = 0$

$sin(z) = \frac{e^{iz}-e{-iz}}{2i} = 0$

$e^{iz} = e^{-iz}$

$\frac{e^{iz}}{e^{-iz}} = 1$

$e^{2iz} = 1$

$ln(e^{2iz}) = ln(1)$

$2iz = 0$

$z = 0$

However, my problem is z can be $\pi$ or $k\pi$. What should I do?

2) $e^{iz} = -1$

$e^{iz} = i^2$

$ln(e^{iz}) = ln(i^2)$

$iz = 2ln(i)$

$z = \frac{2ln(i)}{i}$

From $z = \frac{2ln(i)}{i}$, is this a final answer?

Thank you.

2. Originally Posted by noppawit
I have 2 problems that I cannot finish/ not sure.

Solve for all values of z such that,
1) $sin(z) = 0$
$sin(z) = \frac{e^{iz}-e{-iz}}{2i} = 0$
$e^{iz} = e^{-iz}$
$\frac{e^{iz}}{e^{-iz}} = 1$
$e^{2iz} = 1$
$ln(e^{2iz}) = ln(1)$
$2iz = 0$
$z = 0$
In my view, you are misusing the ‘log’ function completely.
First of all, the notation for the complex form is $\log$ and not $\ln$.
The definition: $\log(z)=\ln(|z|)+i\arg(z)$.
Many of the usual properties do not hold.
$\log(i)= i\left( {\frac{\pi }{2} + 2n\pi } \right)$ this means that $\log \left( {i^2 } \right) \ne 2\log (i)$.

But this is true: $\log \left( {e^z } \right) = z + 2n\pi i$.