I'm studying complex analysis and I'm wondering about all complex values of $z$ that satisfy the equation:
$$ \cosh(z)=0 \,\, . $$
Is there a smart way to show all values that vanish with the equation above? If yes, how could I demonstrate this? What are these values?
Thank you!
$$\cosh(z)=0\Longleftrightarrow$$ $$\frac{e^z+e^{-z}}{2}=0\Longleftrightarrow$$ $$e^z+e^{-z}=0\Longleftrightarrow$$ $$e^{-z}\left(1+e^{2z}\right)=0\Longleftrightarrow$$
Since $e^z$ is never zero for any $z\in\mathbb{C}$, no solution exists for $e^{-z}=0$:
$$1+e^{2z}=0\Longleftrightarrow$$ $$e^{2z}=-1\Longleftrightarrow$$ $$2z=i\pi(2n+1)\Longleftrightarrow$$ $$z=\frac{i\pi(2n+1)}{2}$$
With $n\in\mathbb{Z}$
Hint
$$\frac {e^z+e^{-z}}{2}=0 \iff e^{2z}=-1$$
Added
Let $z=x+iy$. Let $w:=e^{2z}$ then $w=e^{2z}=e^{2x+2yi}=e^{2x}(\cos2y+i\sin2y)$.
Now, just compare radius and argument with $-1$.
We have $|w|=e^{2x}=1=|-1|$ so $x=0$.
Lastly, \begin{align} \arg(w)&=2y\\ \arg(-1)&=\pi \end{align}
Then, $2y=\pi+2k\pi\rightarrow y=\frac \pi 2+k\pi$.
So we get the solutions: $z=0+\left(\frac{\pi}2+k\pi\right) i$.
