Complex Taylor Series Circles of Convergence I am trying to find the Taylor Series and circles of convergence for three different functions.
i) $\frac{\sin{z}}{z}$ which I determined the Taylor series to be $\sum_{n=0}^\infty (-1)^n\frac{z^{2n}}{(2n+1)!}$
ii) $z\cosh{z^2}$ becomes $\sum_{n=0}^\infty \frac{z(z^2)^{2n}}{2n!}$
iii)$\frac{z}{z^4+9}$ becomes $\sum_{n=0}^\infty (-1)^n\frac{z^{4n+1}}{9^{n+1}}$
I am stumped on how to find the circles of convergence.  I understand that the circle of convergence is defined by the radius of convergence, but I having problems determining the radius of convergence.
 A: Here's a sketch of the ratio test, and a worked example: If $(a_{n})$ is a sequence of real or complex terms (possibly containing variables), and if
$$
\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = L < 1,
$$
(in the sense that the indicated limit exists, and is smaller than $1$), then $\sum\limits_{n=0}^{\infty} |a_{n}|$ converges, so $\sum\limits_{n=0}^{\infty} a_{n}$ converges.
For example, suppose we want the circle of convergence of the power series
$$
\sum_{n=0}^{\infty} (-1)^{n} \frac{(3z - 5)^{2n+1}}{n^{2}},
\quad\text{so that}\quad
a_{n} = (-1)^{n} \frac{(3z - 5)^{2n+1}}{n^{2}}.
$$
We calculate (using $|-1| = 1$, and omitting several steps of algebra)
$$
\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|
  = \lim_{n \to \infty} \left|\frac{(3z - 5)^{2n+3}}{(n + 1)^{2}}
                        \cdot \frac{n^{2}}{(3z - 5)^{2n+1}}\right|
  = |3z - 5|^{2} \lim_{n \to \infty} \frac{n^{2}}{(n + 1)^{2}}
  = |3z - 5|^{2}
  = L.
$$
Consequently, the original series converges if $|3z - 5| < 1$ (taking square roots preserves the inequality), i.e., if $|z - 5/3| < 1/3$, and diverges if $|z - 5/3| > 1/3$.
The radius of convergence is $1/3$, and the center is $z_{0} = 5/3$.
The three examples at hand work similarly (though you'll have to cancel some ratios of factorials to find $L$). If you find that $L = 0$ independently of $z$, that series converges absolutely for all complex $z$.
