Proving that a complex power series is differentiable

Asked Oct 16 '18 at 11:16

Active Oct 16 '18 at 11:16

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Let $p(z)=a_0+\dots+a_nz^n, n\geqslant 1, a_n\neq 0.$ I want to show that $f:\mathbb{R}\rightarrow\mathbb{R}$ is a complex differentiable function, where $f=1/p(z)$. I already showed that there is an $R>0$ such that $\left|z\right|>R$ implies that $\left|z^nf(z)\right|<2/\left|a_n\right|$.

Any hints to get me going?

asked Oct 16 '18 at 11:16

David W.

Hint: https://math.stackexchange.com/questions/2957694 – Nosrati Oct 16 '18 at 12:40
I don't see what the hint is... – David W. Oct 16 '18 at 13:09
What means that a function $\mathbb{R}\rightarrow\mathbb{R}$ is complex differentiable? – Martín-Blas Pérez Pinilla Oct 16 '18 at 20:11
What if $p(z) = z?$ – zhw. Oct 17 '18 at 02:14
You want to show $f$ is entire? How? Wouldn't that be contradicting the Fundamental Theorem of Algebra? – Shashi Oct 18 '18 at 09:08

Proving that a complex power series is differentiable

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