Geometric representation of all the points for $\cos(z) = 2$, $z \in \mathbb C$ So I am having trouble visualizing the solutions to
$$\cos(z) = 2, z \in \mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $\cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?
If $z=x+iy$ then

$$ \cos(x+iy) = \cos(x)\cos(iy) - \sin(x)\sin(iy) = \cos(x)\cosh(y) - i\sin(x)\sinh(y) $$

Therefore

\begin{align} \cos(x)\cosh(y) &amp;= 2 \\ \sin(x)\sinh(y) &amp;= 0 \end{align}

Obviously, no real solution exists, so $y\ne 0 \implies \sin x = 0$. 

But $\cosh (y) &gt; 0, \forall y \implies \cos x &gt; 0 \implies \cos(x) = 1 \implies x = 2n\pi$, which leaves

$$ \cosh(y) = 2 \implies y = \pm \cosh^{-1}(2) = \pm \ln (2 + \sqrt{3}) $$

So the solution is

$$ x = 2n\pi \pm i\ln(2 + \sqrt{3}) $$

For a visualization, think of the solutions as points on the 2 lines $y = \pm \ln(2+\sqrt{3})$, spaced out evenly by $2n\pi$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $\cos z = \text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $\ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.