How do I go about finding the set of points in $\Bbb C$ where $f(z)=(z^4 + z)e^{\bar{z}}$ is differentiable $$f(z) = {(z^4 + z)}e^{\bar{z}}$$
In the exercise there are $6$ possible answers:
$\{{0,-1, e^{i\pi/3}, e^{-i\pi/3}}\}$ ; $\{{0,1, e^{i\pi/3}, e^{-i\pi/3}}\}$ ; $\{{0,-1, e^{2i\pi/3}, e^{-2i\pi/3}}\}$ ; $\{{0,1, e^{2i\pi/3}, e^{-2i\pi/3}}\}$ ; $\{0\}$ ; $\emptyset$.
The correct answer according to the exam solutions appears to be:
$${0,-1, e^{i\pi/3}, e^{-i\pi/3}}$$
At first I tried to go down the route of applying the Cauchy-Riemann equations, however it proved to be very difficult and it appears it can't simply be solved that way, however I can be wrong.
Since, if a function is analytic in a region R, it can be written as a power series in there, and $f(0)=f(1)=f(e^{i\pi})=f(e^{-i\pi})=0$ , while the same doesn't happen in the other points where if I expand it into the form of a series it can't be expressed in the form of a power series I think, then that would eliminate answer $2$, $3$ and $4$.
If what I'm saying is correct then do I need to check if the Cauchy-Riemann equations are satisfied at these points? The process is very length and demanding, also doesn't look like the way to go. I assume there is something I might be missing.
 A: Notice that $f$ is differentiable at a point $z_0\in\mathbb C$ if and only if it is real differentiable (which your function is everywhere) and satisfies the Cauchy--Riemann equations at that point, which is
$\partial_{\bar z} f(z_0) = 0$.
Since $g(z) = z^4+z$ is holomorphic, it follows by the product rile and $\partial_{\bar z} g(z)=0$ that $$\partial_{\bar z} f(z) =g (z)\cdot \partial_{\bar z} e^{\bar z} = f(z)$$
i.e. $f(z)$ satisfies the Cauchy--Riemann equations exactly when $f(z_0)=0$, and since $e^{\bar z}$ has no zeros, you are left with solving $z^4+z=0$, giving the first solution you mention.

Here $\partial_{\bar z} = \frac 1 2 \left(\partial_x + i \partial_y\right)$ is one of the Wirtinger derivatives. You can check by hand by multiplying things out that $\partial_{\bar z} (u+iv) = 0$ is equivalent to the Cauchy--Riemann equations,
and that $\partial_{\bar z}$ and that this derivative and $\partial_{z}= \frac 1 2 \left(\partial_x - i \partial_y\right)$ satisfy the product rule and commute, for example.
