I saw on another post that if $f(z)$ is entire, and $\lim\frac{f(z)}{z}=0$, then $g(z)=\frac{f(z)-f(0)}{z}$ is entire. Why is this true? I understand that by the way it's defined, $g$ is continuous (if you add on the point $f'(0)$ for when $z=0$).
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1Because $z=0$ is a removable singularity. $\lim_{z \to 0} (z-0)g(z) = 0$. – copper.hat Mar 12 '20 at 17:37
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Because if$$f(z)=a_0+a_1z+a_2z^2+\cdots,\tag1$$then$$\frac{f(z)-f(0)}z=a_1+a_2z+a_3z^2+\cdots\tag2$$and, since the series $(1)$ converges everywhere, then so does the series $(2)$. So, its sum is an entire function.
José Carlos Santos
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Oh, so when the comment says it is entire "by addition and quotient of entire functions", this was not the real reason? As in, quotients of entire functions in general are not entire. (Like $1/z$) – Mar 12 '20 at 17:36
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Yes, something about the zeros of those functions must be added. – José Carlos Santos Mar 12 '20 at 17:38
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Wait, how exactly do we know that the second series also converges? Edit: oh nevermind, I think I understand... since it's for all $z$. – Mar 12 '20 at 17:40
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Oh nevermind, it's because they literally have the same radius of convergence. Thanks! – Mar 12 '20 at 17:56