The modulus of the conjugate of z = frac{-2+i | vedclass

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(B) Given $z = \frac{-2+i}{(1-2i)^2}$.First,simplify the denominator: $(1-2i)^2 = 1^2 + (2i)^2 - 2(1)(2i) = 1 - 4 - 4i = -3 - 4i$.So,$z = \frac{-2+i}{

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$\frac{1}{\sqrt{5}}$

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(B) Given $z = \frac{-2+i}{(1-2i)^2}$.First,simplify the denominator: $(1-2i)^2 = 1^2 + (2i)^2 - 2(1)(2i) = 1 - 4 - 4i = -3 - 4i$.So,$z = \frac{-2+i}{-3-4i} = \frac{2-i}{3+4i}$.Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:$z = \frac{(2-i)(3-4i)}{(3+4i)(3-4i)} = \frac{6 - 8i - 3i + 4i^2}{3^2 + 4^2} = \frac{6 - 11i - 4}{9 + 16} = \frac{2 - 11i}{25}$.The conjugate is $\bar{z} = \frac{2 + 11i}{25}$.The modulus is $|\bar{z}| = |z| = \sqrt{(\frac{2}{25})^2 + (-\frac{11}{25})^2} = \sqrt{\frac{4 + 121}{625}} = \sqrt{\frac{125}{625}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

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

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