solving $\cos z + \sin z = i$ I found a solution but when testing it, it does not work....?  
$$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi$$  
Upon testing it in W|A with $k=0$, I don't get the required result $(=i)$. I'm not sure where I went wrong and I did this a few times already.

You may have applied the quadratic form incorrectly as I can see.
You have:
$$e^{2iz}(1+i)+2e^{iz}-1+i=0$$
with $a=1+i,b=2,c=-1+i$.
$$\implies e^{iz}=\frac{-2\pm \sqrt{2^2-4(i+1)(i-1)}}{2(1+i)}=\frac{-2\pm\sqrt{4-4(i^2-1)}}{2(i+1)}=\frac{-2\pm\sqrt{4-4(-2)}}{2(i+1)}$$
$$=\frac{-2\pm\sqrt{4+8}}{2(i+1)}=\frac{-2\pm\sqrt{12}}{2(i+1)}=\frac{-2\pm2\sqrt{3}}{2(i+1)}=\frac{-1\pm\sqrt{3}}{1+i}$$.
$$\therefore iz=\ln\left(\frac{-1\pm\sqrt{3}}{1+i}\right)\Rightarrow z=-i\ln\left(\frac{-1\pm\sqrt{3}}{1+i}\right)+2k\pi$$
which I believe can be further simplified with complex algebra.
