What is an example of a complex function with a zero of infinite order (other than the zero map $z \to 0$)?

Question

I can't find any example other than the zero map which may very well be because there isn't any other.

From Taylor's theorem, it seems obvious that the only possible example is the zero map, since the only power series with only zero coefficients is zero. Is this correct?

(The order of a zero is defined here.)

@GerryMyerson Yes, $f$ is analytic. I've just included the definition for the order of a zero. — Stephen, Dec 23 '19 at 16:03
OK. The Taylor series of an analytic function converges to the function, so I think your argument is OK. — Gerry Myerson, Dec 23 '19 at 16:06

score 5 · Accepted Answer · answered Dec 23 '19 at 16:09

5

By the identity theorem, if two analytic functions $f$ and $g$ satisfy $f^{(n)}(z_0)=g^{(n)}(z_0)$ for all $n\in\mathbb{N}_0$, then $f=g$ everywhere on $\mathbb{C}$. Hence, if $f$ has a zero of infinite order somewhere in $\mathbb{C}$, then $f\equiv 0$, since $f^{(n)}(z_0)=0=g^{(n)}(z_0)$ for all $n\in\mathbb{N}_0$, taking $g$ to be the zero map.

answered Dec 23 '19 at 16:09

csch2

4,557

I think the identity theorem could be perhaps too strong as one might need this result to prove it. I think that Gerry Myerson's comment suffices though. – Maths Wizzard May 30 '23 at 16:08

What is an example of a complex function with a zero of infinite order (other than the zero map $z \to 0$)?

1 Answers1