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)


Note that 
$$\cos(\arctan(z)) + \sin(\arctan(z)) = \frac{1}{\sqrt{1+z^2}} + \frac{z}{\sqrt{1+z^2}}.$$
So, if you can solve the equation 
$$\frac{1}{\sqrt{1+z^2}} + \frac{z}{\sqrt{1+z^2}} = 2,$$
(which is quadratic) then you can solve the original.
