I would like to show that both equations have infinitely many solutions in $\mathbb C.$
In the first case, I tried to use Picard's theorem similarly as in here. That is if function $f(z):=\cos\frac1z-\frac1z$ has an essential singularity at zero, then $\cos(z)-z$ has an essential singularity at $\infty$ and the rest is Picard's cannon. But is that really the case? Of course, $\lim_{z\to0} f(z)$ does not exists but $\lim_{z\to0}1/f(z)=0$ if I'm not mistaken, and by definition it is not essential.
As for the second equation, the situation is worse since function $\cos z-\bar z$ is not even differentiable, although I still believe it has infinitely many roots.
In both cases, I tried to use expansion $\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y,$ compare real and imaginary parts and eventually eliminate one of the variables. But the resulting equation is quite messy and I was not able to conclude much.
Best Answer
It is not so messy. The two equations being $$\cos (a) \cosh (b)-a=0 \tag 1$$ $$\sin (a) \sinh (b)+b=0 \tag 2$$ From $(1)$ $$b=\cosh ^{-1}(a \sec (a))\tag 3$$ Plug it in $(2)$ to obtain (assuming $a>1$) $$\tan (a)\sqrt{a^2-\cos ^2(a)} +\cosh ^{-1}(a \sec (a))=0\tag 4$$ For the analysis, it could better (asuming $\cos(a)\neq 0$) to write it as a function $$f(a)=\sin (a)\sqrt{a^2-\cos ^2(a)} +\cosh ^{-1}(a \sec (a))\cos(a)\tag 5$$ which, for sure, has a lot of discontinuities but, as usual, an infinite number of roots which are closer and closer to $2n\pi$. Using series expansion $$f(a)=\cosh ^{-1}(2 \pi n)+\frac{4 \pi ^2 n^2 }{\sqrt{4 \pi ^2 n^2-1}}(a-2 n\pi )+O\left((a-2 n\pi)^2\right)$$ and a first estimate $$\color{red}{a_0=2 n\pi-\frac{\sqrt{4 \pi ^2 n^2-1} }{4 \pi ^2 n^2}\cosh ^{-1}(2 \pi n)}\sim 2 n\pi- \frac{\log (4 \pi n)}{2 \pi n}$$ which is quite accurate $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 5.88650 & 5.86956 \\ 2 & 12.3107 & 12.3086 \\ 3 & 18.6573 & 18.6567 \\ 4 & 24.9770 & 24.9768 \\ 5 & 31.2842 & 31.2841 \\ 6 & 37.5845 & 37.5845 \\ \end{array} \right)$$