radius of convergence of $1/(1+z^2)$ about $z=2$ using geometric series approach

Question

I would like to calculate the radius of convergence of $f(z)= 1/(1+z^2)$ about $z=2$ using the geometric series approach.

Let me first state that according to a theorem, the radius of convergence about a point $z_o$ is the smallest distance from $z_o$ to a singularity.

Using this method, the distances from $z=2$ to the singularities $z=i$ and $z=-i$ are $\sqrt{5}$ which is the radius of convergence. However, I get a different answer using the geometric series approach.

I can re-write f(z) as $(1/5)/(1+(4/5)(z-2)+(1/5)(z-2)^2)$. Comparing this to the expression $1/(1+z)$, we want $|(4/5)(z-2)+(1/5)(z-2)^2|<1$ for convergence. This results in $|z|<3$ implying the radius of convergence is $3-2=1$.

Why is the answer here different?

thanks

Answer 1

The geometric series approach does not work naively here to find the radius of convergence. You are correct that the radius of convergence of the series $$\frac{1}{5}\sum_{n=0}^\infty \left(-\frac{4}{5}(z-2)-\frac{1}{5}(z-2)^2\right)^n$$ around $z=2$ is $1$ (here by "radius of convergence" I mean the radius of the largest disk around $2$ on which it converges--but this disk is actually not the entire set on which the series converges). However, the partial sums of this series are not the same as the partial sums of the Taylor series for $f(z)$ around $z=2$. Indeed, notice that each term of the geometric series, when expanded, contributes to several different terms of the Taylor series, since it will involve several different powers of $z-2$. Each term of the Taylor series is actually a combination of pieces from several different terms of the geometric series. So if you cut off the geometric series at a certain point, you have all of the contributions to some of the terms of the Taylor series, but also some partial contributions to later terms of the Taylor series.

So since the two series do not have the same partial sums, there is no obvious reason they should converge for the same values of $z$. And in this case, they indeed do not. It turns out that when you group together the terms from the geometric series according to the power of $z-2$ they contain, some magic cancellation happens which makes the radius of convergence increase from $1$ to $\sqrt{5}$.

In fact, the set on which the geometric series converges is not even a disk centered at $2$ at all--it is the funny-shaped region $|z^2-4|<5$. You can compare the regions of convergence of the Taylor series and the geometric series at http://www.wolframalpha.com/input/?i=implicit+plot+(x%5E2-y%5E2-4)%5E2%2B(2xy)%5E2%3D25+and++(x-2)%5E2%2By%5E2%3D5. The nice circle is the region where the Taylor series converges, and the weird lumpy thing is the region where the geometric series converges.

Answer 2

Your strategy looks exactly right to me, but somewhere you went off the rails. Let me make typing and notation easier by setting $z-2=t$, $z=t+2$. Then the mathematics becomes much clearer. You have $$ \frac1{1+z^2}=\frac1{1+(t+2)^2}=\frac15\biggl[\frac1{1+\frac45t+\frac15t^2}\biggr]\,, $$ and I agree totally. But then, using Quadratic Formula, you should factor the denominator of the fraction in brackets, $1+\frac45t+\frac15t^2=(1-\alpha t)(1-\bar\alpha t)$, with $\alpha=\frac{-2+i}5$, a complex quantity of absolute value $1\big/\sqrt5$. Without bothering to do the actual computation, you know that $$ \frac1{(1-\alpha t)(1-\bar\alpha t)}=\frac A{1-\alpha t}+\frac B{1-\bar\alpha t}\,. $$ This says that the radii of convergence of the two series are both $|1/\alpha|=|1/\bar\alpha|=\sqrt5$, and I think this is the way that it was hoped that you would get the answer.

Notice that we don’t care what $A$ and $B$ are, but if I’ve not gone astray, $A=\bar B=(3-4i)/5$. This is enough for you to see that when you add the two complex series together term by term, you get a real series, as of course you must.