I'm looking through a past paper question and I'm stuck on this one:
Let $f(z)=\frac{10z}{z^2+z-6}$.
- What is the radius of convergence for the Taylor series of $f$ about the point $z=i$?
I've separated $f(z)$ into partial fractions $\frac{10z}{z^2+z-6}=\frac{6}{z+3}+\frac{4}{z-2}$ and now wish to look at the Taylor expansion about $z=i$, I write $z=i+w$ then
$\frac{6}{w+i+3}+\frac{4}{w+i-2}$?
Is this right so far? I'm struggling, if someone could provide assistance/solutions for me to analyse that would be so helpful!
In my Complex Analysis Course with Professor Barry Hughes at the University of Melbourne, we were taught a theorem that is stated as follows:
If the power series $\sum\limits_{n=0}^{\infty}c_n(z-z_0)^n$ diverges for $z=z_1$, then it necessarily diverges wherever $\lvert z-z_0 \rvert >\lvert z_1-z_0\rvert$.
Similarly, if the power series $\sum\limits_{n=0}^{\infty}c_n(z-z_0)^n$ converges for $z=z_1$, then it converges absolutely in the open disc $\lvert z-z_0 \rvert <\lvert z_1-z_0\rvert$.
So now let's apply this:
We need only find the closest point $z_1$ to $z_0=i$ such that the Taylor series does not converge at $z_1$ - in this case, it will be one of the poles of $f(z)$.
By separating the fraction into $f(z)=\frac{6}{z+3}+\frac{4}{z-2}$, you've located the poles of $f(z)$. You can see that as $z$ approaches the points $z_1=2,\ z_2=-3$, $f(z)$ will attain arbitrarily large values and so the Taylor series will diverge. So we can apply the theorem and state that since the Taylor series cannot possibly converge for $z=z_1$ and $z=z_2$, we know that the radius of convergence $r$ is less than $min(\lvert z_0-z_1\rvert,\lvert z_0-z_2\rvert)$. Now, all that is left to show is that the Taylor Series will converge for all $\lvert z-z_0 \rvert<min(\lvert z_0-z_1\rvert,\lvert z_0-z_2\rvert)$. (Hint: To do this, write down the Taylor coefficients $c_n$ and see if you can show that the infinite sum converges)