Find all complex solutions to an exponential equation Find all complex numbers z such that $e^{-2iz}/4 + e^{-iz}/2 + 1 + 2e^{iz} + 4e^{2iz} = 0 $
Rewriting the left-hand side using Eulers formula doesn't seem to get me anywhere. Need some help with this one!
Thanks in advance!
 A: Let $w=2 e^{i z}$.  Then
$$w^2+w+1+\frac{1}{w}+\frac{1}{w^2}=0$$
This can be transformed into
$$\left ( w+\frac{1}{w}\right)^2 + w+\frac{1}{w} -1=0$$
So let $y=w+1/w$ and solve the resulting quadratic:
$$y^2+y-1=0 \implies y= \frac{-1\pm\sqrt{5}}{2} = \phi_{\pm}$$
Now solve
$$w^2-\phi_{\pm} w + 1=0$$
which implies that
$$w = \frac{\phi_{\pm} \pm \sqrt{\phi_{\pm}^2-4}}{2} = \frac{\phi_{\pm}}{2} \pm i \sqrt{1-\left ( \frac{\phi_{\pm}}{2}\right)^2}$$
Also, you may notice that
$$\frac{\phi_+}{2} = \cos{\frac{2 \pi}{5}}$$
$$\frac{\phi_-}{2} = \cos{\frac{4\pi}{5}}$$
So you can put $w$ into the form $e^{i \theta}$ and solve for $z$ accordingly.  We get 4 solutions for $w$:
$$w_1 = e^{i 2 \pi/5}  = 2 e^{i z_1}$$
$$w_2 = e^{-i 2 \pi/5}= 2 e^{i z_2}$$
$$w_3 = e^{i 4 \pi/5}= 2 e^{i z_3}$$
$$w_4 = e^{-i 4 \pi/5}= 2 e^{i z_4}$$
Solve for the $z$'s using logarithms.  For example,
$$z_1 = \frac{2 \pi}{5} + 2 \pi k_1 + i \log{2}$$
for any $k_1 \in \mathbb{Z}$.  You can do the rest similarly.
A: This might get you started.  Multiply through by $e^{2iz}$, obtaining
$4e^{4iz} + 2e^{3iz} + e^{2iz} + \frac{1}{2}e^{iz} + \frac{1}{4} = 0$,
then set $w = e^{iz}$, obtaining
$4w^4 + 2w^3 + w^2 + \frac{1}{2}w + \frac{1}{4} =  0$,
multiply by $4$ to make things look nice:
$16w^4 + 8w^3 + 4w^2 + 2w + 1 = 0$.
Now you can solve it as a quartic (from the well -known procedure) and work backwards from there to get $z$ such that $e^{iz} = w$!
