Question

The conjugate complex number of $\frac{2-i}{(1-2i{)}^2}$ is

• A. $\frac{2}{25}+\frac{11}{25}i$
• B. $\frac{2}{25}-\frac{11}{25}i$
• C. $-\frac{2}{25}+\frac{11}{25}i$
• D. $-\frac{2}{25}-\frac{11}{25}i$

Answer 1

The correct option is D $-\frac{2}{25}-\frac{11}{25}i$

Rationalizing the denominator
$\frac{2-i}{(1-2i{)}^2}\times \frac{(1+2i{)}^2}{(1+2i{)}^2}$

$=\frac{(1+2i{)}^2(2-i)}{[1^2-(2i{)}^2{]}^2}$

$=\frac{(1-4+4i)(2-i)}{(1+4{)}^2}$

$\therefore \frac{(-3+4i)(2-i)}{(5{)}^2}$

$=\frac{-2+11i}{25}$

Complex conjugate of $a+ib$ is $a-ib$

Hence, the conjugate of $\frac{-2+11i}{25}=\frac{-2-11i}{25}$