Suppose $$f(z)= \frac{\sin(z)}{z^2 +64}.$$ I want to argue that the radius of convergence of the power series expansion of $$f$$ at $$z=6$$ is $$10.$$ Since the function $$f$$ is holomorphic inside the open ball $$\mathbb{B}(6, 10$$) ( Open ball centered at 6 with radius 8), the function must be analytic on this region. Moreover, $$10$$ is the largest possible radius of such balls centered at $$z=6.$$ Is this argument correct? Thank you.
$$f$$ is analytic except for poles at $$\pm8i$$. Since $$|6\pm8i|=10$$, this shows that $$f$$ is analytic in $$B(6,10)$$ but not in $$B(6,r)$$ for $$r>10$$; hence the radius of convergence of the power series is $$10$$.
Your approach is correct. Another reasoning is that the power series of $$\frac{\sin z}{z^2+64}$$ converges at a neighborhood of $$z=z_0$$ if the power series of $$\sin z$$, $$\frac{1}{z+8i}$$ and $$\frac{1}{z-8i}$$ converge at that neighborhood. Since the radius of convergence of $$\sin z$$ is $$\infty$$ at any $$z=z_0$$, we only investigate $$\frac{1}{z+8i}$$ and $$\frac{1}{z-8i}$$. Note that $${\frac{1}{z+8i}=\sum_{n=0}^\infty \frac{(-1)^n}{(6+8i)^n}(z-6)^n \\ \frac{1}{z-8i}=\sum_{n=0}^\infty \frac{(-1)^n}{(6-8i)^n}(z-6)^n. }$$ Since the radius of convergence of both the above series is $$10$$ and $$\frac{\sin z}{z^2+64}$$ diverges for $$|z-6|=10+\epsilon$$ for some arbitrarily small value of $$\epsilon$$ due to the divergence of $$\frac{1}{z-8i}$$ and $$\frac{1}{z+8i}$$, then so is our total radius of convergence.
• "the power series of $\frac{\sin z}{z^2+64}$ converges at $z=z_0$ if and only if the power series of $\sin z$, $\frac{1}{z+8i}$ and $\frac{1}{z-8i}$ converge at $z=z_0$" really? Why is that? – David C. Ullrich May 7 at 15:15