Question

Write (a+ib/a-ib)^2-(a-ib/a+ib)^2 in the form x +iy please reply me this

Answer 1

$\textbf{Given:}$

$(\dfrac{a+i\,b}{a-i\,b})^2-(\dfrac{a-i\,b}{a+i\,b})^2$

$\textbf{To find:}$

$\text{x+iy form of (\dfrac{a+i\,b}{a-i\,b})^2-(\dfrac{a-i\,b}{a+i\,b})^2 }$

$\textbf{Solution:}$

$\text{Consider}$

$(\dfrac{a+i\,b}{a-i\,b})^2$

$=\dfrac{(a+i\,b)^2}{(a-i\,b)^2}$

$=\dfrac{a^2-b^2+i\,2ab}{a^2-b^2-i\,2ab}$

$(\dfrac{a-i\,b}{a+i\,b})^2$

$=\dfrac{(a-i\,b)^2}{(a+i\,b)^2}$

$=\dfrac{a^2-b^2-i\,2ab}{a^2-b^2+i\,2ab}$

$\text{Now,}$

$(\dfrac{a+i\,b}{a-i\,b})^2-(\dfrac{a-i\,b}{a+i\,b})^2$

$=(\dfrac{a^2-b^2+i\,2ab}{a^2-b^2-i\,2ab})-(\dfrac{a^2-b^2-i\,2ab}{a^2-b^2+i\,2ab})$

$=\dfrac{(a^2-b^2+i\,2ab)^2-(a^2-b^2-i\,2ab)^2}{(a^2-b^2-i\,2ab)(a^2-b^2+i\,2ab)}$

$=\dfrac{((a^2-b^2)^2-4a^2b^2+i\,4ab(a^2-b^2))-((a^2-b^2)^2-4a^2b^2-i\,4ab(a^2-b^2))}{(a^2-b^2)^2+\,4a^2b^2}$

$=\dfrac{(a^2-b^2)^2-4a^2b^2+i\,4ab(a^2-b^2)-(a^2-b^2)^2+4a^2b^2+i\,4ab(a^2-b^2)}{(a^2-b^2)^2+\,4a^2b^2}$

$=\dfrac{i\,8ab(a^2-b^2)}{(a^2-b^2)^2+\,4a^2b^2}$

$=0+i\,\dfrac{8ab(a^2-b^2)}{(a^2-b^2)^2+\,4a^2b^2}$

$\therefore\bf\,(\dfrac{a+i\,b}{a-i\,b})^2-(\dfrac{a-i\,b}{a+i\,b})^2=0+i\,\dfrac{8ab(a^2-b^2)}{(a^2-b^2)^2+\,4a^2b^2}$