$\cos(z)=z$ and $\cos(z)=\bar z$ 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.
 A: 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
$$\left(
\begin{array}{ccc}
n & \text{estimate} & \text{solution} \\
 1 & 5.88650 &  5.86956 \\
 2 & 12.3107 &  12.3086 \\
 3 & 18.6573 &  18.6567 \\
 4 & 24.9770 &  24.9768 \\
 5 & 31.2842 &  31.2841 \\
 6 & 37.5845 &  37.5845 \\
\end{array}
\right)$$
