Convergence Problem Does there exist a sequence of polynomials $P_{n}(z)$ such that $e^{P_{n}(z)}$
converges uniformly on compact sets to $z?$ 
Any chance that theorems like Montel's should be used?
 A: You can also use Rouche's theorem: if you did have uniform convergence, then for large enough $n$ on the circle $|z| = 1$ you will have ${\displaystyle |e^{P_n(z)} - z| < |z| = 1}$. Hence by Rouche, $e^{P_n(z)}$ and $z$ will have the same number of zeroes inside the circle. But exponentials are never zero, while $z$ has a zero inside the circle, a contradiction.
A: Here are two different methods to answer this:


*

*Apply Hurwitz's theorem to the unit disk.

*If such $(P_n)$ exists, then $(e^{-P_n(z)})$ converges uniformly to $\frac{1}{z}$ on the unit circle.  Find the integral of each of these functions on the circle.

A: Argument Principle: Consider the contour integral over the unit circle.  Apply the argument principle.  Since $e^z$ is never zero, the change in argument will be zero.  Since $z$ has one zero, the change in argument around the circle will be $2\pi$.  Hence by IVT there is a point where the arguments differ by $\pi$ so that $z$ and $e^{P_n(z)}$ point in opposite directions.
A: I think you get the idea from the other answers that in general this can't be done. However, if your domain is contained in a disk that does not contain the origin, then you can approximate the power series for
$$
\log(z_0) + \log\left(1+\frac{z-z_0}{z_0}\right)=\log(z_0)+\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\left(\frac{z-z_0}{z_0}\right)^k
$$
which converges for $|z-z_0|<|z_0|$ and
$$
e^{\log(z_0) + \log\left(1+\frac{z-z_0}{z_0}\right)} = z
$$
