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