If $f,g$ are entire functions, $p,q$ are polynomials in the complex plane satisfying $\exp(f)+p=\exp(g)+q, f(0)=g(0)$ prove that $f=g$ and $p=q$?
Here. My approach is very much messy and calculative. I am lost. How to proceed with this one?
I was trying in the way that if $q\ne p,$ then the polynomial $q-p$ is of a finite degree and therefore I calculate the derivative for $\exp(f)-\exp(g),$ but this seems a bad approach and I also can't express my solution here due to the very much intuition and vague calculation.
I expect a neat and crucial approach can anyone help.
Best Answer
I got one solution. We have to use little Picard theorem see $e^f-e^g=q-p$ ,now $e^f-e^g=e^g(e^{f-g}-1).$ Now see that if $f-g$ is not identically zero, then as it is entire non constant (as $f(0)=g(0)$ given), by little Picard theorem the range of $(f-g)(z)$ omits at most one point so..if $f(z)-g(z)$ can take $2n\pi i$ for infinitely many $n$..so $e^(f-g)-1$ is zero for infinitely many $z$ ..so in the RHS $q-p$ has infinitely many roots and, as they are polynomial, so $q-p=0$ ,so $q=p$ identically so this implies $g-f\in\{2n\pi:n\in\Bbb Z\}$ and by continuity $g-f$ is constant and as $g(0)=f(0)$ so $g=f$ identically on $C$..