# Math Help - complex trig Q

1. ## complex trig Q

Hi, I am not seeing the following step:

We are given $cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2}$. Therefore,

$e^{2iz}-e^{iz}+1=0$, and sp by the quadratic formula....

I don't see how the got $cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2}$ into the form $e^{2iz}-e^{iz}+1=0$. I can see the mult. the 2 and subtracted $e^{iz}$ over, but I don't see how the got the $e^{2iz}$ term.

Thanks

2. In the original equation, multiply both sides by 2, bring the 1 to the LHS, and multiply both sides by $e^{iz}$. Can you see what happens?

3. $e^{iz}+e^{-iz}-1=0$
$e^{iz}e^{iz}+e^{iz}e^{-iz}-e^{iz}=0$
$e^{iz+iz}+e^{-iz+iz}-e^{iz}=0$
$e^{2iz}+e^{0}-e^{iz}=0$
$e^{2iz}-e^{iz}+1=0$