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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.

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$f$ 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 $10$.

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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.

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    $\begingroup$ "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? $\endgroup$ – David C. Ullrich May 7 at 15:15
  • $\begingroup$ @DavidC.Ullrich, you are right. I hope you do not mind to revise my updated answer. $\endgroup$ – Mostafa Ayaz May 7 at 15:53

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