# Write the conjugate of

Question:

Write the conjugate of $\frac{2-i}{(1-2 i)^{2}}$.

Solution:

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

$=\frac{2-i}{1-4-4 i}$

$=\frac{2-i}{-3-4 i}$

$=\frac{-2+i}{3+4 i}$

$=\frac{i-2}{3+4 i} \times \frac{3-4 i}{3-4 i}$

$=\frac{3 i-4 i^{2}-6+8 i}{3^{2}-4^{2} i^{2}}$

$=\frac{11 i+4-6}{9+16}$

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

$\therefore$ Conjugate of $\frac{2-i}{(1-2 i)^{2}}=\left(-\frac{2}{25}+\frac{11}{25} i\right)=-\frac{2}{25}-\frac{11}{25} i$

Hence, Conjugate of $\frac{2-i}{(1-2 i)^{2}}$ is $-\frac{2}{25}-\frac{11}{25} i$.