Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$

Question

Use the definition $$ \cos z =\frac{e^{iz}+e^{-iz}}{2} $$ to find $2$ imaginary numbers having a cosine of $4$.

I tried two approaches, both of which ended in failure:

$$ 8=e^{iz}+e^{-iz}\\ \ln8=0\\ \\ \space \\ 8=\cos z+i\sin z+\cos z-i\sin z\\ z = \arccos4 $$

What approach should I take instead?

Answer 1

You're using $\log$ incorrectly, even in the real sense. It's not true, for instance, that $\log\left(e+e^{-1}\right)=0$.

Hint: Instead, multiply both sides of $8=e^{iz}+e^{-iz}$ by $e^{iz}$ and solve the resulting quadratic in $e^{iz}$.

Answer 2

$u=e^{ix}$
${{u+{1\over{u}}}\over2}=4$
$u+{1\over{u}}=8$
$u^2+1=8u$
$u^2-8u+1=0$
$u=e^{ix}={{8\pm\sqrt{60}}\over2} ={{8\pm{2\sqrt{15}}}\over2} ={4\pm\sqrt{15}}$
$ix=\ln(4\pm\sqrt{15})$
$x={\ln(4\pm\sqrt{15})\over{i}} =-i\ln(4\pm\sqrt{15})$