I'm working with the function
$w(z) = \frac{1}{z-3}+\frac{1}{z-4}\ \mathrm{with}\ z_0=1$.
I found the Taylor Series to be
$$-\sum_{n=0}^{\infty}\Big(\frac{1}{2}\Big)^{n+1}(z-1)^n-\sum_{m=0}^{\infty}\Big(\frac{1}{3}\Big)^{n+1}(z-1)^n.$$
Now I'm asked to find the region of convergence, which is defined as
$\Big|\frac{z-z_0}{a-z_0}\Big|<1$.
How do I do this when there are two singularities?
Best Answer
The Taylor series for an analytic function converges in the largest disc where the function is analytic. Therefore, if $z_0=1$, then the series will converge in a disc which is the minimum of the distance from $z_0$ to $3$ and $z_0$ to 4. So the radius of convergence is $|1-3| = 2$. You should draw it. In general the series at $z_0$ converges inside the largest disc centered at $z_0$ which doesn't contain any singularities of the function. The radius of that circle is the radius of convergence.
Sometimes when adding series, it might happen that singularities cancel each other out, but in general when adding series, the radius of convergence is the smallest of the two you are adding. But for example: $$ 1 = \frac{1-z}{1-z} = \frac{1}{1-z} + \frac{-z}{1-z} = \sum_{n=0}^\infty z^n + \sum_{k=1}^\infty (-1)z^{k} = 1 $$ So even though the two series in the middle had radius of convergence 1, the function itself (a constant) has radius of convergence $\infty$.