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$f$ is an entire function such that $|zf(z)-1+e^z|\leq 1+|z| \ \forall z\in \mathbb C$ then which of the folowing are true?

(1) $f'(0)=1$,

(2) $f'(0)=-1/2$,

(3) $f'(0)=-1/3$,

(4) $f'(0)= -1/4 $

My approach:

Since $f$ is an entire function so it has a power series expansion then $f(z)=\sum_{n=0}^{\infty}a_nz^n$.

So, $|z\sum_{n=0}^{\infty}a_nz^n-1+e^z|=|z\sum_{n=0}^{\infty}a_nz^n+\sum_{n=1}^{\infty}\frac{z^n}{n!}|=|a_0z+\sum_{n=1}^{\infty}(a_nz+\frac{1}{n!})z^n|$

So, $|a_0z+\sum_{n=1}^{\infty}(a_nz+\frac{1}{n!})z^n|\leq 1+|z|$

Now I can not understand how to proceed. Can anyone please help me? Thank you in advance.

Gary
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Aritra
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    Can you show that $zf(z)-1+e^z$ must be a linear function? Then note that $zf(z)-1+e^z$ is $$ zf(0) + z^2 f'(0) + O(z^3 ) - 1 + 1 + z + \frac{{z^2 }}{2} + O(z^3 ) = z(f(0) + 1) + \left( {f'(0) + \frac{1}{2}} \right)z^2 + O(z^3 ) $$ as $z\to 0$. – Gary May 19 '22 at 07:08
  • @Gary from the last inequality we get $f(z)$ is linear. Is not? Like $f(z)$ is of the form $(f(0)+1)z$. – Aritra May 19 '22 at 07:16
  • First you show that $zf(z)-1+e^z$ is linear ($f(z)$ is not linear). Then use the expansion I gave to compute $f'(0)$. – Gary May 19 '22 at 07:24

1 Answers1

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Let $g(z)=zf(z)-1+e^z$. Then you always have $|g(z)|\leqslant1+|z|$, and therefore $g(z)=az+b$ for some numbers $a,b\in\Bbb C$. But\begin{align}zf(z)-1+e^z=az+b&\iff zf(z)=az+b+1-e^z\\&\iff f(z)=a+\frac{b+1-e^z}z.\end{align}Since $f$ is entire, it has no pole at $0$, and so $b+1-e^0=0$; in other words, $b=0$. So,$$z\ne0\implies f(z)=a-\frac{e^z-1}z.$$But then$$f(z)=a-1-\frac z2-\frac{z^2}6+\cdots,$$and therefore $f'(0)=-\frac12$.

José Carlos Santos
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