Show that the function $g(z) = f(e ^z )$ is not a polynomial.

Let $f :\mathbb C \rightarrow \mathbb C$ be an entire function.

Show that the function $g(z) = f(e ^z )$ is not a polynomial.

What is the technique to show a function is a polynomial?

One common way to show that functions cannot be (non-constant) polynomials is that (non-constant) polynomials only have finitely many roots.

If $g(z)$ were polynomial, then $g(z) = g(0)$ would have finitely many solutions. However, $g(2\pi i k) = f(e^{2\pi i k}) = f(1) = g(0)$ holds true for all $k\in \mathbb{Z}$.

• Seems we don't need the hypothesis of f being entire. – gary Apr 9 '15 at 7:19
• @learnmore: I hink he is referring to the value $g(0)=f(1)$ , where $f(1)$ is seen as a complex constant. – gary Apr 9 '15 at 7:23
• thanks for the answer; i got it – Learnmore Apr 9 '15 at 7:34

Assume $f(e^z)$ is a polynomial. Then so is $f'(e^z)$ and its derivative :

$f'(e^z)e^z$ . But $e^z (a_0+a_1z+...+a_nz^n)$ is not a polynomial. If it was,

( a non-constant) polynomial, $e^z =\frac {p(z)}{q(z)}$ would have singularities, or $e^z$ would be a polynomial itself. But, "borrowing" (ripping of) Rolf's answer, $e^z$ is not a polynomial since it has no zeros.