The answer is \(\frac{5}{78}+ \frac{25}{78}i\)
\(\frac{5}{3-15i} =\frac{5}{3-15i}* \frac{3+15i}{3+15i} = \frac{5(3+15i)}{(3-15i)(3+15i)} \\ \\
a^{2} -b^{2} =(a+b)(a-b) \\ \\ 4 +
\frac{5(3+15i)}{(3-15i)(3+15i)} = \frac{15+75i}{ 3^{2}-(15i)^{2} } = \frac{15+75i}{9-15^{2}i^{2}} \\ \\ i^{2}=-1 \\ \\
\frac{15+75i}{9-15^{2}i^{2}} =\frac{15+75i}{9-225*(-1)} =\frac{15+75i}{9+225} = \frac{15+75i}{234} = \frac{3*5+3*25i}{3*78} = \frac{3(5+25i)}{3*78} = \\ \\
= \frac{5+25i}{78}= \frac{5}{78}+ \frac{25i}{78} =\frac{5}{78}+ \frac{25}{78}i\)
