The answer is [tex]\frac{12+8i}{13} [/tex]
[tex] \frac{4}{3-2i} = \frac{4(3+2i)}{(3-2i)(3+2i)} \\ \\ (a-b)(a+b) = a^{2} -b^{2} \\ \\
\frac{4(3+2i)}{(3-2i)(3+2i)} = \frac{4*3+4*2i}{3 ^{2}- (2i)^{2} }= \frac{12+8i}{9- (2)^{2}(i)^{2} }= \frac{12+8i}{9-4(i)^{2}} \\ \\
(i)^{2} = -1 \\ \\
\frac{12+8i}{9-4(i)^{2}}== \frac{12+8i}{9-4*(-1)}}= \frac{12+8i}{9+4} = \frac{12+8i}{13} [/tex]
