Find all the complex roots of $\cosh(z)=\frac{1}{2}$ Find all the complex roots of $\cosh(z)=\frac{1}{2}$
I have so far used the formula $\cosh(z)=\frac{e^z+e^{-z}}{2}$ to get:
\begin{align} & \frac{e^z+e^{-z}}{2}=\frac{1}{2} \\
\implies & e^z+e^{-z}=1
\end{align}
Now, I'm not sure where to go. 
 A: You can rewrite the expression with a common denominator, since $e^{-z} = \dfrac{1}{e^{z}}$:
$$\dfrac{(e^{z})^{2}}{e^{z}}+\dfrac{1}{e^{z}}=1$$
Multiply by $e^{z}$ on both sides:
$$(e^{z})^{2} + 1 = e^{z}$$
Then subtract both sides by $e^{z}$:
$$(e^{z})^{2} - e^{z} + 1 = 0$$
Solve via the quadratic formula:
$$e^{z} = \dfrac{1\pm\sqrt{-3}}{2}$$
$$z = \ln\left(\dfrac{1\pm i\sqrt{3}}{2}\right)$$
And taking the natural log of a complex number $z$ is $\ln{|z|} + i\arg{z}$, giving:
$$z = \pm\;i\dfrac{\pi}{3}\pm2k\pi$$
A: The method you are following is pretty standard, here are (the beginnings of) two others.


*

*$\cosh z=\frac12$, so $\sinh^2z=\cosh^2z-1=-\frac34$, so $\sinh z=\pm i\frac{\sqrt3}2$, so
$$e^z=\cosh z+\sinh z=\frac{1\pm i\sqrt3}2\ .$$

*$\cos(iz)=\cosh z=\frac12$, so
$$iz=\pm\frac\pi3\pm 2k\pi\ .$$


I'll also continue your argument a bit - wasn't going to as it was in someone else's answer, but that answer has now been deleted (don't know why, there was nothing wrong with it).


*

*Multiply both sides of $e^z+e^{-z}=1$ by $e^z$ and rearrange to get
$$(e^z)^2-(e^z)+1=0$$
then solve the quadratic.


I'll leave you to finish the details.
A: $\displaystyle e^z+e^{-z}=1 \tag 1$
$\displaystyle z=x+iy \tag 2$
From $1$ and $2$ you can say that:
$\displaystyle e^z+e^{-z}=e^{x+iy}+e^{-x-iy}=e^{x}(e^{iy})+e^{-x}(e^{-iy})=e^x(cos(y)+isin(y))+e^{-x}(cos(y)-isin(y))=\cos(y)(e^x+e^{-x})+i\sin(y)(e^x-e^{-x})$
$\displaystyle=2\cos(y)\cosh(x)+2i\sin(y)\sinh(x)=1+0i$
$\displaystyle \to \sin(y)\sinh(x)=0$ and $\displaystyle \cos(y)\cosh(x)=\dfrac12$
$\sinh(x)=0, \cos(y)=\dfrac12$, So:

$\displaystyle x=0$ and $\displaystyle y=2k\pi\pm \pi/3,$ or simply: $\displaystyle z=2k\pi\pm \pi/3$

