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)


Now here is a purely algebraic solution. As you have noted, the problem is equivalent to $$(1 + i){e^{iz}} - (1 - i){e^{ - iz}} = 4i$$consider the new variable $x$ as ${x=e^{iz}}$in terms of the newly defined variable, the problem reads as$$(1 + i)x - {{(1 - i)} \over x} = 4i$$multiplying by $x$, we get $$(1 + i){x^2} - 4ix - (1 - i) = 0$$which is a standard second order equation. The solutions are $$x = {{4i \pm \sqrt { - 16 + 4(1 - i)(1 + i)} } \over {2(1 + i)}} = \left( {1 \pm {{\sqrt 2 } \over 2}} \right)(1 + i)$$so that $$z = {1 \over i}\ln x =  - i\left( {\ln \left( {1 \pm {{\sqrt 2 } \over 2}} \right) + \ln (1 + i)} \right)$$which is hopefully what you are looking for (note that a more careful analysis in the logarithm part yields solutions in higher Riemann planes).
