Question
What is the conjugate of $\frac{2-\text{i}}{(1-2\text{i})^2}$

Answer

We have $\text{z}=\frac{2-\text{i}}{(1-2\text{i})^2}$
$\text{z}=\frac{2-\text{i}}{1+4\text{i}^2-4\text{i}}=\frac{2-\text{i}}{1-4-4\text{i}}=\frac{2-\text{i}}{-3-4\text{i}}$
$=\frac{(2-\text{i})}{-(3+4\text{i})}=-\Big[\frac{(2-\text{i})(3-4)}{(3+4\text{i})(3-4\text{i})}\Big]$
$=-\Big(\frac{6-8\text{i}-3\text{i}+4\text{i}^2}{9+16}\Big)=-\frac{(-11\text{i}+2)}{25}$
$=\frac{-1}{25}(2-11\text{i})=\frac{1}{25}(-2+11\text{i})$
$\therefore\ \bar{\text{z}}=\frac{1}{25}(-2-11\text{i})=\frac{-2}{25}-\frac{11}{25}\text{i}$

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