# solve complex conjugation equation

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)

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$z+\overline z=2\operatorname{Re}z$. –  Marc van Leeuwen Oct 12 '12 at 7:07

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.

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