Why are we ignoring the singularity point $z=0$ in that solution?

Asked Jul 27 '21 at 14:30

Active Jul 27 '21 at 15:29

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Let $f$ be entire s.t $|f(z)| \leq |e^z -1|$, prove that there exist $c \in \mathbb{C}$ s.t $$f(z) = c \cdot (e^z -1)$$

So the idea here is to show that $h(z) = \frac{f(z)}{e^z -1}$ is entire and than by Liouville's theorem it is constant. the solution says that for every $z \neq 2i \pi n$ for natural $n$ the function is holomorphic and is equal to constant and we get what we asked for (this part is understood for me) in the case of $z = 2 i\pi n$ we get that $$|f(2 i\pi n)| \leq |e^{2i \pi n} -1| = 0$$ and hence $z= 2 i\pi n$ is removable singularity.

My question is what about $z=0$ why we are ignoring that point?

edited Jul 27 '21 at 15:29

Bio

asked Jul 27 '21 at 14:30

Sagigever

1,428

All points $z=2i \pi n$ are removable singularities, including $z=0$ ($n=0$). – Martin R Jul 27 '21 at 14:34
I do not understand why, it seems that $z=0$ the limit is divergence – Sagigever Jul 27 '21 at 14:41
Since $|f(z)| \leq |e^z-1|$, wherever $e^z-1$ has a zero, $f$ does too. – Ian Jul 27 '21 at 14:46
1

Note that your conclusion $f(z) = c \cdot e^z -1$, is wrong, it should be $f(z) = c \cdot (e^z -1)$. – Martin R Jul 27 '21 at 14:50
Ok now I understand, thank you very much @MartinR yes it was typo, i fixed that – Sagigever Jul 27 '21 at 14:56

Why are we ignoring the singularity point $z=0$ in that solution?

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