Question

The real and imaginary parts of $\frac{a+ib}{a-ib}$ are:

• A. $a^2-b^2,2ab$
• B. $\frac{a^2+b^2}{a^2-b^2},\frac{2ab}{a^2-b^2}$
• C. $\frac{a^2-b^2}{a^2+b^2},\frac{2ab}{a^2+b^2}$
• D. $\frac{a^2+b^2}{a^2-b^2},\frac{2ab}{a^2+b^2}$

Answer 1

Answer: (C)

$\frac{a^2-b^2}{a^2+b^2},\frac{2ab}{a^2+b^2}$

$z=\frac{a+ib}{a-ib}$

Multiply & divide by $(a+ib)$ so as to make the denominator real.

$\therefore \frac{(a+ib)}{(a-ib)}\times \frac{(a+ib)}{(a+ib)}$

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

$=\frac{a^2-b^2+2abi}{a^2+b^2}$

$=\left(\frac{a^2-b^2}{a^2+b^2}\right)+\left(\frac{2ab}{a^2+b^2}\right)i$

Imaginary part $=\frac{2ab}{a^2+b^2}$