If frac{(p + i)^2}{2p - i} = mu + ilambda,the | vedclass

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(D) Given $\mu + i\lambda = \frac{(p + i)^2}{2p - i}$.Taking the modulus on both sides,we get $|\mu + i\lambda| = \left|\frac{(p + i)^2}{2p - i}\right

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$\frac{(p^2 + 1)^2}{4p^2 + 1}$

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(D) Given $\mu + i\lambda = \frac{(p + i)^2}{2p - i}$.Taking the modulus on both sides,we get $|\mu + i\lambda| = \left|\frac{(p + i)^2}{2p - i}\right|$.Since $|z_1/z_2| = |z_1|/|z_2|$ and $|z^n| = |z|^n$,we have $\sqrt{\mu^2 + \lambda^2} = \frac{|p + i|^2}{|2p - i|}$.Calculating the moduli: $|p + i| = \sqrt{p^2 + 1}$ and $|2p - i| = \sqrt{(2p)^2 + (-1)^2} = \sqrt{4p^2 + 1}$.Thus,$\sqrt{\mu^2 + \lambda^2} = \frac{(\sqrt{p^2 + 1})^2}{\sqrt{4p^2 + 1}} = \frac{p^2 + 1}{\sqrt{4p^2 + 1}}$.Squaring both sides,we get $\mu^2 + \lambda^2 = \frac{(p^2 + 1)^2}{4p^2 + 1}$.

If $\frac{(p + i)^2}{2p - i} = \mu + i\lambda$,then $\mu^2 + \lambda^2$ is equal to

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