Answer
$\frac{3i}{(4-5i)^2}=-\frac{120}{1681}-\frac{27}{1681}i$
Work Step by Step
$(4-5i)^2=4^2-2(4)(5i)+(5i)^2=16-40i-25=-9-40i$
$\frac{3i}{(4-5i)^2}=\frac{3i}{-9-40i}=\frac{3i}{-9-40i}\frac{-9+40i}{-9+40i}=\frac{3i(-9+40i)}{(-9-40i)(-9+40i)}=\frac{-27i+120i^2}{81-1600i^2}=\frac{-27i-120}{81+1600}=-\frac{120}{1681}-\frac{27}{1681}i$
