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Find real part of $\left(\frac{a+bi}{a-bi}\right)^2-\left(\frac{a-bi}{a+bi}\right)^2$, for any $a,b$ (they are real numbers)

Please help guys

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    $z+\overline z=2\operatorname{Re}z$.2012-10-12

2 Answers 2

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HINT: Plug in some numbers, calculate $\overline{\left(\frac{a+bi}{a-bi}\right)^2}$ and think about Marc's comment.

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Let $A+iB=(a+ib)^2,$ so $A-iB=(a-ib)^2$

$\left(\frac{a+bi}{a-bi}\right)^2-\left(\frac{a-bi}{a+bi}\right)^2$

$=\frac{A+iB}{A-iB}-\frac{A-iB}{A+iB}$

$=\frac{(A+iB)^2-(A-iB)^2}{A^2+B^2}$

$=\frac{4ABi}{A^2+B^2}$

So the number is purely imaginary.