How to show $\frac{e^{z^{2}}}{z^{1995}}$ is analytic?

Question

For $\frac{e^{z^{2}}}{z^{1995}}$, I am thinking if I can rewrite it in form of $a+bi$ and then apply Cauchy-Schwartz equations.

Although I can do it for $e^{z^2}$, I am not sure about what to do with $z^{1995}$ on the denominator.

Really appreciate for helping.

Answer 1

Hint: Put $z=r.e^{i\theta}$ and use CR-equations in polar form. Something like this

$f(r.e^{i\theta})=u(r,\theta)+iv(r,\theta)$

Then CR-equations will be $r.u_r=v_\theta, u_\theta=-r.v_r$