Why is not $e^{x}e^{-iy}$ not complex differentiable nor holomorphic? I'm struggling a bit with finding out why $e^{x}e^{-iy}$ is neither complex differentiable or holomorphic. Would someone mind explaining why? In the past i've learnt that the exponential function is atleast complex differentiable at some point.
I've checked if the derivitives satisfies Cauchy-Riemann equations, which gets me that it satisfies one the two. But i dont really know how to find out if there is any way that
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
is equal at some point.
$u(x,y)=e^x\cos(-y)$
$\frac{\partial u}{\partial x}=e^{x}\cos(-y)$
$\frac{\partial u}{\partial y}=e^{x}\sin(-y)$
$v(x,y)=e^x\sin(-y)$
$\frac{\partial v}{\partial x}=e^{x}\sin(-y)$
$\frac{\partial v}{\partial y}=-e^{x}\cos(-y)$
Thanks!
 A: Complex differentiable is the same as holomorphic.
The exponential function is holomorphic at every point, but the function $z\mapsto\overline z$ is antiholomorphic.
That is why your composite function $z\mapsto e^{\overline z}$ is (at every point) antiholomorphic and not holomorphic.
A: You calculated partial derivatives in Cauchy-Riemann equations. Let's check when the equations are satisfied:
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \iff e^x \cos(-y) = -e^x\cos(-y)\iff \cos(-y) = 0$$
$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \iff e^x \sin(-y) = -e^x\sin(-y)\iff \sin(-y) = 0$$
For which $y\in\mathbb R$ is both sine and cosine of $-y$ zero?
A: To find the set of differentiability (or analyticity possibly), just look for the set of values of $z$ which satisfy $\partial f/\partial\overline z=0.$ In your case, $\frac{\partial }{\partial\overline z}(e^{\overline z})=e^{\overline z}=0$ holds for no values of $z$, so the set is empty. 
You may like to see a similar question  here.
