Solve $\cos{z}+\sin{z}=2$ I am trying to solve the question:


$\cos{z}+\sin{z}=2$
Where $z \in \mathbb{C}$


I think I know how to solve $\cos{z}+\sin{z}=-1$:
$1+2\cos^2{\frac{z}{2}}-1+2\sin \frac{z}{2}\cos{\frac{z}{2}}=0\\
2\cos{\frac{z}{2}}(\cos{\frac{z}{2}}+\sin{\frac{z}{2}})=0$ 
etc... (that is, if the double angle identity holds true when the 'angle' is a complex number - I might be wrong about this)
My other methods involve: 


*

*trying to substitute $\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$ and $\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$. This seems to be the most obvious method, but I can't work out the next step after $$e^{iz}-e^{-iz}+(e^{iz}+e^{-iz})i=4i$$

*substituting $2=2(\sin^2{z}+\cos^2{z})$

*substituting $\sin{z}=\cos(\frac{\pi}{2}-z)$ (again, not really sure if this can be done)


I think you were on the right track. Putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$e^{iz}-e^{-iz}+(e^{iz}+e^{-iz})i=4i\implies e^{-y+ix}-e^{y-ix}+\left(e^{-y+ix}+e^{y-ix}\right)i=4i\iff$$
$$e^y\left(\cos x-i\sin x+i(\cos x-i\sin x)\right)+e^{-y}\left(\cos x+i\sin x+i(\cos x+i\sin x)\right)=4i\iff$$
$$e^y\left(\cos x+\sin x\right)+e^{-y}\left(\cos x-\sin x\right)+\left[e^y\left(\cos x-\sin x\right)+e^{-y}\left(\cos x+\sin x\right)\right]i=4i$$
And thus you get two equations:
$$\begin{align*}e^y\left(\cos x+\sin x\right)+e^{-y}\left(\cos x-\sin x\right)=0\\e^y\left(\cos x-\sin x\right)+e^{-y}\left(\cos x+\sin x\right)=4\end{align*}$$
If you add\substract both equations you get
$$\cos x\left(e^y+e^{-y}\right)=2\iff\cos x\cosh x=1$$
$$\sin x\left(e^y-e^{-y}\right)=-2\iff \sin x\sinh x=-1$$
You now have two transcendental, real equations to solve
