Solve $\cos(z)+\sin(z)=2$
My work: \begin{alignat}{2} \cos(z)+\sin(z)=2 & \implies e^{iz}&&+e^{-iz}+e^{-iz}-e^{iz}=4\\ &\implies e^{-iz}&&=2 \\ &\implies e^{-iz}&&=e^{\ln(2)+2\pi ni} \\ &\implies-iz&&=\ln(2)+2\pi ni \\ &\implies z&&=i\ln(2)-2 \pi n \\ \end{alignat} but apparently the answer is $z=\frac{\pi}{4}+2n\pi \pm i\ln(\sqrt{2}+1)$.
Where have I made a mistake and why?
Hint: $$\cos{(z)}=\frac{e^{iz}+e^{-iz}}{2}$$ $$\sin{(z)}=\frac{e^{iz}-e^{-iz}}{2i}$$