If a=|bar{a}| and b=|bar{b}|,then left(frac{b | vedclass

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(B) Given that $a = |\bar{a}|$ and $b = |\bar{b}|$. Since $|\bar{a}| = |a|$,we have $a = |a|$ and $b = |b|$.We know that for any complex number $z$,$z

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$\left(\frac{\bar{a}-\bar{b}}{a b}\right)^2$

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(B) Given that $a = |\bar{a}|$ and $b = |\bar{b}|$. Since $|\bar{a}| = |a|$,we have $a = |a|$ and $b = |b|$.We know that for any complex number $z$,$z \bar{z} = |z|^2$. Thus,$\bar{a} = \frac{|a|^2}{a} = \frac{a^2}{a} = a$ is not necessarily true,but rather $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{a \bar{a} \cdot a} = \frac{1}{a \bar{a}} \cdot \bar{a} = \frac{1}{a}$.Actually,using $z \bar{z} = |z|^2$,we have $\frac{\bar{a}}{a^2} = \frac{\bar{a}}{|a|^2} = \frac{\bar{a}}{a \bar{a}} = \frac{1}{a}$.Similarly,$\frac{\bar{b}}{b^2} = \frac{1}{b}$.Therefore,$\left(\frac{\bar{a}}{a^2} - \frac{\bar{b}}{b^2}\right)^2 = \left(\frac{1}{a} - \frac{1}{b}\right)^2 = \left(\frac{b-a}{ab}\right)^2 = \left(\frac{a-b}{ab}\right)^2$.

If $a=|\bar{a}|$ and $b=|\bar{b}|$,then $\left(\frac{\bar{a}}{a^2}-\frac{\bar{b}}{b^2}\right)^2=$

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