Solve the following equations:
$$ \cos\left(z\right)=2+4i$$ $$ \text{Log}\left(z\right)=\left(1+i\right)\pi$$
$$ \arctan\left(z\right)=1+i$$
For the first one we know that $$cos(z)=\cos\left(z\right)=\cos\left(x\right)\cosh\left(y\right)-i\sin\left(x\right)\sinh\left(y\right)=2+4i$$
Which implies $$\cos\left(x\right)\cosh\left(y\right)=2$$ and $$\sin\left(x\right)\sinh\left(y\right)=-4$$
Then how to proceed?
For the second one $$\text{Log}\left(z\right)=\ln\left|z\right|+i\text{Arg}(z)=\left(1+i\right)\pi$$ Which implies $$\ln\left|z\right|=\pi$$ and $$\text{Arg}(z)=\pi$$
So the solution to the equation is $z=e^\pi e^{\pi(2k+1)}$ for $k \in \mathbb Z$.
For the last one $$\arctan\left(z\right)=\frac{i}{2}\log\left(\frac{i+z}{i-z}\right)=1+i$$
I don't know how to continue
Using you approach for the first ones $$\cos\left(x\right)\cosh\left(y\right)=2\tag1$$ $$\sin\left(x\right)\sinh\left(y\right)=-4\tag2$$
$(1)$gives $\cos(x)=2 \,\text{sech}(y)$ and $(2)$ gives $\sin(x)=-4 \,\text{csch}(y)$
$$\cos^2(x)+\sin^2(x)=\color{red}{16 \,\text{csch}^2(y)+4 \,\text{sech}^2(y)=1} \tag 3$$
The "red" equation does not show obvious roots and its solution requires numerical methods. Graphing, you would notice that the function $$f(x)=\log\Big[16 \,\text{csch}^2(y)+4 \,\text{sech}^2(y) \Big]$$ is very linear as soon as $y>1$ and the solution $y=\pm 2.19857$ is very easy to find. Then, reusing $(1)$, $\cos(x)=0.43845$ then $x$ (do not forget the modulo).