Is $ze^{z}$ differentiable? As the title says, I am looking at the function: 
$$f(z) = z e^{z},$$
and I want to know whether it is differentiable - and if so, to find its derivative (which I can do). 
Usually I would either make use of the definition from first principles or the Cauchy–Riemann conditions; however, both seem very messy. I further know that $e^{z}$ is entire in the complex plane but I don't know if that is a sufficent condition to prove its differentiability. 
I have seen cases which cite calculating $\frac{\partial f}{\partial \bar{z}}$, but I don't understand the implication or how to achieve this. 
Any help would be greatly appreciated. Thanks!
 A: With $\;z=x+iy\;$ :
$$f(z)=(x+iy)(e^x\cos y+ie^x\sin y)=\overbrace{xe^x\cos y-ye^x\sin y}^{=u(x,y)}+\left(\overbrace{xe^x\sin y+ye^x\cos y}^{=v(x,y)}\right)i$$
Now Cauchy-Riemann:
$$u'_x=e^x\left(\cos y+x\cos y-y\sin y\right)=v'_y$$
$$u_y=e^x\left(-x\sin y-\sin y-y\cos y\right)=-v'_x$$
A: Note that $f(z)=ze^z$ is not a function of $\bar z$.  Therefore, we have
$$\frac{\partial f(z)}{\partial \bar z}=0 \tag 1$$
The expression in $(1)$ is equivalent to the Cauchy-Riemann equations
$$\begin{align}\frac{\partial u(x,y)}{\partial x}&=\frac{\partial v(x,y)}{\partial y} \tag {2a}\\\\
\frac{\partial u(x,y)}{\partial y}&=-\frac{\partial v(x,y)}{\partial x}\tag {2b}\end{align}$$
We show the equivalence of $(1)$ with $(2a)$ and $(2b)$ by writing $f(z)=u(x,y)+iv(x,y)$.  Then, using $x=\frac12(z+\bar z)$ and $y=\frac1{2i}(z-\bar z)$ along with the chain rule reveals
$$\begin{align}\frac{\partial f(z)}{\partial \bar z}&=\frac12 \frac{\partial f(z)}{\partial x} +\frac{i}{2}\frac{\partial f(z)}{\partial y}\\\\
&=\frac12\left(\frac{\partial u(x,y)}{\partial x}-\frac{\partial u(x,y)}{\partial y}\right)+i\frac12\left(\frac{\partial u(x,y)}{\partial y}+\frac{\partial v(x,y)}{\partial x}\right) \tag 3\\\\
\end{align}$$
Setting $(3)$ to $0$ establishes the coveted equivalence!  And since $f$ satisfies the Cauchy-Riemann equations, it is differentiable.
