Solve $\sin z - \cos z = \mathrm{i}$ Where $z$ is a complex number. 
$\sin z-\cos z = \mathrm{i} \implies \frac{\mathrm{e}^{\mathrm{i}z}-\mathrm{e}^{-\mathrm{i}z}}{2\mathrm{i}}-\frac{\mathrm{e}^{\mathrm{i}z}-\mathrm{e}^{-\mathrm{i}z}}{2} = \mathrm{i}$
multiply both sides by $4\mathrm{i}$
$2\mathrm{e}^{\mathrm{i}z}-2\mathrm{e}^{-\mathrm{i}z}-2\mathrm{i}\mathrm{e}^{\mathrm{i}z}-2\mathrm{i}\mathrm{e}^{\mathrm{i}z} = -4$
divide both sides by $2$, bring everything to the LHS
$\Rightarrow \mathrm{e}^{\mathrm{i}z}-\mathrm{e}^{-\mathrm{i}z}-\mathrm{i}\mathrm{e}^{\mathrm{i}z}-\mathrm{i}\mathrm{e}^{\mathrm{i}z}+2 = 0$
multiply by $\mathrm{e}^{\mathrm{i}z}$
$(1-\mathrm{i})\mathrm{e}^{2\mathrm{i}z}+2\mathrm{e}^{\mathrm{i}z}-(\mathrm{i}+1) = 0$.
quadratic formula,
$\mathrm{e}^{\mathrm{i}z} = \frac{-2 \pm \sqrt{12}}{2(1-\mathrm{i})} = \frac{(-1 \pm \sqrt{3})(1+\mathrm{i})}{2}$ 
If the above expression is correct, 
then eventually my final answer becomes $z = \frac{\pi}{4} +2k\pi - \mathrm{i}\ln\left( \frac{-\sqrt{2}+\sqrt{6}}{2} \right)$. Could anyone verify that my final answer for $z$ is correct?
thanks : ) 
 A: Well, we have:


*

*$$\cos\left(x\right)=\frac{\exp\left(xi\right)+\exp\left(-xi\right)}{2}\tag1$$

*$$\sin\left(x\right)=\frac{\exp\left(xi\right)-\exp\left(-xi\right)}{2i}\tag21$$
And we want to solve:
$$\sin\left(\text{z}\right)-\cos\left(\text{z}\right)=i\tag3$$
So we can also write:
$$\frac{\exp\left(\text{z}i\right)-\exp\left(-\text{z}i\right)}{2i}-\frac{\exp\left(\text{z}i\right)+\exp\left(-\text{z}i\right)}{2}=i\tag4$$
Substitute $\text{y}:=\exp\left(\text{z}i\right)$, so we get:
$$\frac{\text{y}-\frac{1}{\text{y}}}{2i}-\frac{\text{y}+\frac{1}{\text{y}}}{2}=i\space\Longleftrightarrow\space$$
$$\text{y}=\left(\frac{1}{2}+\frac{i}{2}\right)\cdot\left(\sqrt{3}-1\right)\space\vee\space\text{y}=\left(-\frac{1}{2}-\frac{i}{2}\right)\cdot\left(1+\sqrt{3}\right)\tag5$$
A: You calculation is ok, but you can simplify a bit the final answer to $z_0=\frac{\pi}4+2k\pi+\frac i2\ln(2+\sqrt{3})$
Also you are missing some solutions, notice that if $z_0$ is solution then $z_1=-\frac{\pi}2-z_0$ is solution too.
Indeed $\cos(z_1)-\sin(z_1)=-\sin(z_0)--\cos(z_0)=\cos(z_0)-\sin(z_0)$.
$z_1=-\frac{3\pi}4+2k\pi-\frac i2\ln(2+\sqrt{3})$
A: Your answer is correct. Here another shortcut solution:


*

*$\sin z - \cos z = i \stackrel{\cos \frac{\pi}{4} =\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}}{\Longrightarrow} \sin (z - \frac{\pi}{4}) =\frac{\sqrt{2}}{2}i$

*$\arcsin w = -i\log (iw + \sqrt{1-w^2})$
It follows
\begin{eqnarray*} z 
& = & \frac{\pi}{4}  -i \log \left(-\frac{\sqrt{2}}{2} + \sqrt{1+\frac{1}{2}}\right)\\
& = & \frac{\pi}{4} -i \left(\ln \frac{-\sqrt{2}+ \sqrt{6}}{2}+2k\pi i \right) \: \: (k \in \mathbb{N})
\\
& = & \frac{\pi}{4} + 2k\pi - i \ln \frac{\sqrt{6} -\sqrt{2}}{2} \: \: (k \in \mathbb{N})
\end{eqnarray*}
