Question

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

Answer 1

Explanation:

$$\begin{gathered} \frac{\left(2-i\right)}{{\left(1-2i\right)}^2}=\frac{\left(2-i\right)}{\left(1-4i-4\right)} \\ =\frac{\left(2-i\right)}{\left(-3-4i\right)} \\ =\frac{\left(-2+i\right)}{\left(3+4i\right)} \end{gathered}$$

Multiply numerator and denominator with $\left(3-4i\right)$

$$\begin{gathered} \frac{\left(-2+i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}=\frac{\left(-6+3i+8i+4\right)}{\left(9+16\right)} \\ =\frac{\left(-2+11i\right)}{25} \end{gathered}$$

Therefore, Conjugate of $\frac{\left(-2+11i\right)}{25}$is $\frac{\left(-2-11i\right)}{25}$.