Find the solution(s) for $\cos(z+iz)e^{z^2}=0$, where $z$ complex
First, let's look at $e^{z^2}=0$ and assume that there exists such an $z$ so that the equation is satisfied, then
$|e^{z^2}|=0$ which thereby means $e^{\operatorname{Re} z^2}=0$ which is not true as $\exp(x)$ is strictly positive for all real numbers.
Second, and the one I'm having trouble with: $\cos(z+iz)=0$, I used the addition theorem and then substituted it for the exponential function. Eventually I arrive at:
$e^{2iz+2z}=-1$ and I don't know how to continue. Any help would be appreciated.
$e^{z^2}=0$ has no answer but for the second one we have$$z(1+i)=k\pi+\dfrac{\pi}{2}$$or$$z=(k\pi+\dfrac{\pi}{2})\dfrac{1-i}{2}$$