The modulus of the conjugate of $z=\frac{-2+i}{(1-2 i)^2}$ is

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Aicharya · Accepted Answer

$\because z=\frac{-2+i}{(1-2 i)^2}$ $\begin{aligned} & z=\frac{i-2}{1-4-4 i}=\frac{i-2}{-3-4 i}=\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)} \ & z=\frac{2-11 i}{25} 	ext { Then, } \bar{z}=\frac{2+11 i}{25} \ & |\bar{z}|=\left|\frac{2}{25}+\frac{11}{25} iight|=\sqrt{\frac{4}{625}+\frac{121}{625}}=\sqrt{\frac{125}{625}}=\sqrt{\frac{1}{5}} \ & \Rightarrow|\bar{z}|=\frac{1}{\sqrt{5}}\end{aligned}$

The modulus of the conjugate of $z=\frac{-2+i}{(1-2 i)^2}$ is

Solution