# Sequence of polynomials $P_n(z)$ with $e^{P_n(z)}$ converging uniformly on compacta to $z$

I bumped into the following problem, but I'm not sure I know the appropriate way to tackle it.

Is there a sequence of polynomials $P_n(z)$ such that $e^{P_n(z)}$ converges uniformly on compact subsets of $\mathbb{C}$ to $z$?

This seems too good to be true, because if there is such a sequence, then the derivative series also converges uniformly on compacta, giving $|P_n'(z) e^{P_n(z)}-1| \to 0$ for large $n$. So it seems like $e^{P_n(z)}$ would converge to something with poles, unless the polynomials are degree one, which seems like it shouldn't be too bad to find a contradiction in. Would this be a fruitful approach, or is there a slicker way?

• Say $f$ is entire and $f$ has no zero on the circle $|z|=1$. What is $1/(2\pi i)\int_{|z|=1}f'(z)/f(z)\,dz$? – David C. Ullrich Sep 6 '15 at 15:59
• Btw, is "unicompactly" a real word? – David C. Ullrich Sep 6 '15 at 16:00
• Hm, so we could figure out how many roots $P_n'(z)$ has in the unit disk, and then by unicompact convergence, it would be the same as the number of roots of $z$ in the unit disk, i.e., one... so the $P_n$'s are necessarily quadratic? Then fiddle with coefficients? – sourisse Sep 6 '15 at 16:07
• Think about how many roots $e^{P_n}$ has in the disk... – David C. Ullrich Sep 6 '15 at 16:09
• Then you really should modify the title. Using words you invented is not the best way to clue people in on what the question's about... – David C. Ullrich Sep 6 '15 at 16:13

If a sequence $(f_n)$ of entire functions satisfies $f_n(z) \to z$ uniformly on compact subsets of $\mathbb C$, then according to Hurwitz's theorem, there is an $n_0$ such that for $n \ge n_0$ each $f_n$ has exactly one zero in the unit circle. This is surely not possible for $f_n(z) = e^{P_n(z)}$.