Part a).
\begin{align*}
&f(z) = \frac{1}{(z-(3i+4))(z-(-3i + 4))} = \frac{\frac{1}{8 - 6i}}{(3i+4) - z} - \frac{\frac{1}{8 - 6i}}{(4 - 3i) - z}\\
&= \frac{\frac{1}{2} \frac{1}{4-3i} \frac{1}{4+3i}}{1 - \frac{z}{4+3i}} - \frac{\frac{1}{2} \frac{1}{4-3i}^2}{1 - \frac{z}{4-3i}}\\
&= \frac{1}{50} \sum (\frac{z}{4+3i}) ^ n - \frac{1}{2} (\frac{1}{4-3i})^2 \sum (\frac{z}{4-3i})^n
\end{align*}
We need $|\frac{z}{3i+4}| < 1$ and $|\frac{z}{4-3i}| < 1$,
Thus the radius of convergence is $r = 5$, now we consider if the series converges at the $|z | = 5$.
Let $\theta$ be s.t. $\cos \theta = \frac{4}{5}, \sin \theta = \frac{3}{5}$.
Thus, let $t $ be arbitrary,
\begin{align*}
&f(5e^{it}) = K_1 \sum \frac{5 e^{it}}{5e^{i\theta}} ^ n + K_2 \sum \frac{5 e^{it}}{5e^{-i\theta}} \\
&= K_1 \sum e^{ni(t-\theta)} + K_2 \sum e^{ni(t+\theta)}\\
& = K_1 \sum e^{ni(t-\theta)} + |K_2| \sum e^{ni(t+\theta)}e^{i2\theta} \\
& = K_1 \sum e^{ni(t-\theta)} + |K_2| \sum e^{i(nt + (n+2)\theta)}
\end{align*}
if $f(5e^{it})$ converges, then $Re\{f(5e^{it})\}$ converges, since
\begin{align*}
Re\{f(5e^{it})\} = K_1 \sum \cos{n(t-\theta)} + |K_2| \sum \cos {(nt + (n+2)\theta)}
\end{align*}
then $\lim_{n \rightarrow \infty} K_1 \cos{n(t-\theta)} + |K_2| \cos {(nt + (n+2)\theta)} = 0$, since $K_1 = K_2 $, $\lim_{n \rightarrow \infty} K_1 \cos{n(t-\theta)} + |K_2| \cos {(nt + (n+2)\theta)} = \lim_{n \rightarrow \infty} 2K_1 \cos (nt+\theta) \sin (-n\theta - \theta) = 0$, a contradiction.
Thus, on the $|z| = 5$, series does not converge.
Part b).
\begin{align*}
&f(z) = \frac{1}{(z-(3i+4))(z-(-3i + 4))} = \frac{\frac{1}{8 - 6i}}{(3i+4) - z} - \frac{\frac{1}{8 - 6i}}{(4 - 3i) - z}\\
&= \frac{1}{z(6i-8)} \frac{1}{1 - \frac{3i+4}{z}} + \frac{1}{z(8-6i)} \frac{1}{1 - \frac{4-3i}{z}} \\
&= \frac{1}{z(6i-8)} \sum (\frac{3i+4}{z})^n + \frac{1}{z(8-6i)} \sum (\frac{4-3i}{z})^n \\
&= \frac{1}{(6i-8)} \sum \frac{(3i+4)^n}{z^{n+1}} + \frac{1}{(8-6i)} \sum \frac{(4-3i)^n}{z^{n+1}}
\end{align*}
Still, $|\frac{3i+4}{z}| < 1$ and $|\frac{-3i+4}{z}| < 1$ lead to $|z| > 5$. Thus $R = 5$.
Use same approach as above, we can get
\begin{align*}
&f(5e^{it}) = K_3 \sum (\frac{5 e^{i\theta}}{5e^{it}}) ^ n \frac{1}{z} + K_4 \sum (\frac{5 e^{i(-\theta)}}{5e^{it}})^n \frac{1}{z} \\
&= \frac{K_3}{5} \sum e^{ni(\theta - t) - it} + \frac{K_4}{5} \sum e^{ni(-\theta - t) - it} \\
&= \frac{K_3}{5} \sum e^{i(n\theta -(n+1)t)} + \frac{K_4}{5} \sum e^{i(n(-\theta) -(n+1)t)} \\
&= \frac{K_3}{5} \sum e^{i(n\theta -(n+1)t)} + \frac{K_4}{5} \sum e^{i(n(-\theta) -(n+1)t)} \\
&= |\frac{K_3}{5}| \sum e^{i(n\theta -(n+1)t)} e^{i(\theta - \pi)} + |\frac{K_4}{5}| \sum e^{i(n(-\theta) -(n+1)t)} e^{i\theta} \\
&= |\frac{K_3}{5}| \sum e^{i((n+1)\theta -(n+1)t - \pi)} + |\frac{K_4}{5}| \sum e^{i((n+1)(-\theta) -(n+1)t)} \\
\end{align*}
if $f(5e^{it})$ converges, then $Im\{f(5e^{it})\}$ converges, since
\begin{align*}
Im\{f(5e^{it})\} = |K_5| \sum -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sum \sin {((n+1)(-\theta) -(n+1)t)}
\end{align*}
then $\lim_{n \rightarrow \infty} |K_5| -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sin {((n+1)(-\theta) -(n+1)t)} = 0$, since $|K_5| = |K_6|$, $\lim_{n \rightarrow \infty} |K_5| -\sin{((n+1)\theta -(n+1)t)} + |K_6| \sin {((n+1)(-\theta) -(n+1)t)} = -2|K_5| \lim_{n \rightarrow \infty} \cos (-(n+1) t) \sin ((n+1)\theta) = 0$ a contradiction.
Thus, on the $|z| = 5$, series does not converge.
