How do I express this $f(x,y)$ in terms of $f(z)$

Question

$$f(x,y)=e^y\sin x+ie^y\cos x$$

The problem requires me to express in terms of $z$ only.

My attempt:

$$=e^y(\sin x+i\cos x)$$

$$=ie^y(\cos x-i\sin x)$$

If $e^{-i\theta}=(\cos\theta - i\sin \theta)$, then:

$$=ie^ye^{-i\theta}$$

But then I get stuck. I'm not sure what to do with the $e^y$.

Note: - the final answer is given. It says it's:

$$f(z)=ie^{-iz}+c$$

...but I still need to learn how to arrive at that answer.

Answer 1

Write

$$\begin{align} e^y(\sin(x)+i\cos(x))&=i(\cos(x)-i\sin(x))e^y\\\\ &=e^{-ix+y+i\pi/2}\\\\ &=e^{-i(x+iy-\pi/2)}\\\\ &=e^{-i(z-\pi/2)} \end{align}$$