• # question_answer If  $\frac{5(-8+6i)}{{{(1+i)}^{2}}}=a+ib$, then$(a,\,b)$ equals [RPET 1986] A) (15, 20) B) (20, 15) C) $(-15,\,20)$ D) None of these

$\frac{5(-8+6i)}{{{(1+i)}^{2}}}=a+ib$Þ $\frac{-40+30i}{2i}=15+20i=a+ib$ Equating real and imaginary parts, we get $a=15$ and$b=20$.