Find a function $f(z)$ such that its real part is $u(x, y) = e^y\cos(x)$ and it's imaginary part is $v(x, y) = e^y\sin(x)$. Show that for this function Cauchy-Riemann equations are nowhere satisfied.
Basically this question is saying that I need to find a function $f : \mathbb{C} \to \mathbb{C}$ such that $f(z) = f(x +iy) = u(x, y) + iv(x, y)$.
I've shown that for this funtion the Cauchy-Riemann equations are nowhere satisfied.
What I've done for the part I'm having trouble with is the following:
$f(z) = f(x+iy) = u(x, y) + iv(x, y) = e^y\cos(x) + ie^y\sin(x) = e^y(\cos(x) + i\sin(x)) = e^ye^{ix} = e^{y+ix}$
Now I need to somehow manipulate $y + ix$ into the form $x + iy$, how can I go about doing so, so that I can express $e^{y+ix}$ in terms of $z$?
HINT: $y+ix=i(x-iy)=i\overline{z}$.