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Problem Let $p(z)=(z-a_1)(z-a_2)...(z-a_N)$, where $a_1, a_2, ..., a_N$ are distinct complex numbers. Let $M=\min_{1\le{k}\le{N}}|a_k|$. Prove that it is possible to express $\frac{1}{p(z)}$ as a power series $\sum_{n=0}^{\infty}c_nz^n$, for $|z|.

Progress

We look to prove this by induction on $N$. The case $N=1$ is rather simple. We see that $\frac{1}{p(z)}=\frac{1}{z-a_1}=-\frac{1}{a_1}\sum_{n=0}^{\infty}(\frac{z}{a_1})^n$ which converges for $|z| by comparison with $\sum_{n=0}^{\infty}z^n$.

We assume now that the proposition holds for arbitrary $N$ and consider the '$N+1$' case.

Now, $p(z)=(z-a_1)(z-a_2)\cdots(z-a_N)(z-a_{N+1})$. By the assumption of our inductive hypothesis, $\frac{1}{(z-a_1)(z-a_2)\cdots(z-a_N)}$ can be expressed in the form $\sum_{n=0}^{\infty}c_nz^n$ for some complex coefficients $c_n$.

As such, $\frac{1}{p(z)}=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}c_nz^n\sum_{n=0}^{\infty}\frac{1}{(a_{N+1})^n}z^n=-\frac{1}{a_{N+1}}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{c_n}{(a_{N+1})^{n-k}} z^n$ which is of the correct form, but I'm not sure how to demonstrate that convergence holds for $|z|

Any help would be very appreciated. I'm not sure if induction is the best approach to proving this.

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I'll leave previous working here for reference. It seems simply writing $\frac{1}{p(z)}$ as a linear combination of the fractions $\frac{1}{z-a_k}$ for $1\le{k}\le{N}$ is a far less cumbersome approach. Thanks to all who have helped.

Further Problem: Could the radius of convergence exceed $M$?

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    Regarding the "Further Problem": No, it can't, since there's$a$pole at distance $M$ from the origin and the radius of convergence can't extend beyond$a$pole.2011-12-07

2 Answers 2

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Write $f = C \prod_i \frac{1}{1 - z/a_i}$, each has the regular power series expansion for $\frac{1}{1+z}$ for $|z| < |a_i|$, done!

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A standard theorem of complex analysis says: If $z\mapsto f(z)$ is analytic in the disk $D_M:=\{z\in{\mathbb C}\ |\ |z| then the Taylor series of $f$ at the origin has a convergence radius $\rho\geq M$. In your case the function $f(z):={1\over p(z)}$ satisfies the assumption of the theorem, and in addition it has a pole on $\partial D_M$. Therefore one definitely has $\rho=M$.