Finding region where $f(z)=e^{-x}\cos(y) -ie^{-x}\sin(y)$ is analytic

Question

The text I'm using asked me to find on what region $f(z)=e^{-x}\cos(y) -ie^{-x}\sin(y)$ is analytic. So solving the appropriate Cauchy-Riemann Equations. \begin{align} \frac{\partial u}{\partial x}=-e^{-x}\cos(y)=e^{-x}\sin(y)=\frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}=-e^{-x}\sin(y)=-(-e^{-x}\cos(y))=\frac{-\partial v}{\partial x} \end{align} tells me that the region in $\Bbb{C}$ that satisfies those equations consists of complex numbers $z=x+iy$ such that $y=\frac{3\pi}{4}+k\pi$ where $k\in\Bbb{Z}$. But, to me these are points with no neighborhood of differentiable points and makes $f$ nowhere analytic. What am I doing wrong here?

Answer 1

Let $z=x+iy$. $$ e^{-x}\cos(y)-ie^{-x}\sin(y) = e^{-x}(\cos(-y)+i\sin(-y)) = e^{-z} $$ So $f(z)=e^{-z}$, which is entire.