3
$\begingroup$

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?

4 Answers 4

6

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.

  • 0
    Yes, this is nice. It is a common way to prove Hurwitz's theorem.2011-08-21
5

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.
  • 1
    I l$i$ke the second one!2011-08-21
3

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.

1

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 $