Solve cos(z) + sin(z) = i, where z is a complex number and i the imaginary unit So yeah everything is in the title, I tried the trigonometric identity with sin(a+bi) and cos(a+bi) and I tried changing sin(z) and cos(z) for their complex expression, but all to no avail
EDIT: Thanks MPW and David! MPW's trick simplified a lot the expression and then I did like David suggested which had become much easier with this new expression
 A: Hint: You can multiply and divide the left side by $\sqrt 2$ to get 
$$\sqrt 2 \sin\left(z+\frac{\pi}{4}\right)=i.$$
Do you know how to invert the sine function?
Addendum 1: You can generally write
$$a\cos z + b\sin z = \sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos z +\frac{b}{\sqrt{a^2+b^2}}\sin z\right)
$$
and, since in this case, $a=b=1$,
$$\cos z + \sin z = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos z +\frac{1}{\sqrt{2}}\sin z\right)
$$
which we can recognize as
$$\cos z + \sin z = \sqrt{2}\left(\sin{\frac{\pi}{4}}\cos z +\cos{\frac{\pi}{4}}\sin z\right) = \sin\left(\frac{\pi}{4} +z\right)
$$
Addendum 2: We can find a formula for inverting sines as follows. A similar procedure works for all the trig functions. It hinges on your knowing how to invert the exponential function (i.e., using logarithms) and using the quadratic formula (yes, it's valid for complex polynomials). Suppose we are given the complex number $w$ and want to solve for $z$ (I'm using a simplified version of your equation, you can substitute $\hat z = z + \pi/4$).
$$w =\sin z$$ $$w = \frac{e^{iz}-e^{-iz}}{2i}$$ $$\left(2ie^{iz}\right) w = \left(2ie^{iz}\right)\frac{e^{iz}-e^{-iz}}{2i}$$
$$2iw(e^{iz})=(e^{iz})^2 -1$$ $$0 =(e^{iz})^2 -2iw(e^{iz}) - 1$$
This is a quadratic equation (but in $e^{iz }$) of the form $0=az^2+ bz +c$ where $a=1$, $b=-2iw$, and $c=-1$. Use the quadratic formula and simplify to get
$$e^{iz}=(2iw \pm \sqrt{-4w^2+4})/2$$ $$e^{iz}=iw\pm \sqrt{1-w ^2}$$
Now the target is in sight. To solve for $z$, all that remains is to take logarithms and divide by $i$.
A: Try writing it in terms of exponentials:
$$\frac{e^{iz}+e^{-iz}}{2}+\frac{e^{iz}-e^{-iz}}{2i}=i$$
and so
$$i(e^{iz}+e^{-iz})+(e^{iz}-e^{-iz})=-2$$
and multiplying by $e^{iz}$ gives
$$(1+i)(e^{iz})^2-(1-i)=-2(e^{iz})\ .$$
Solve a quadratic to find two possible values for $e^{iz}$, then see if you can solve for $z$.
