Question

The real and imaginary parts of ${\left(\frac{a+ib}{a-ib}\right)}^2-{\left(\frac{a-ib}{a+ib}\right)}^2$ are

• A. $1,\frac{8ab(a^2-b^2)}{(a^2+b^2{)}^2}$
• B. $0,\frac{8ab(a^2+b^2)}{(a^2-b^2{)}^2}$
• C. $0,\frac{8ab(a^2-b^2)}{(a^2+b^2{)}^2}$
• D. $1,\frac{8ab(a^2+b^2)}{(a^2-b^2{)}^2}$

Answer 1

The correct option is C $0,\frac{8ab(a^2-b^2)}{(a^2+b^2{)}^2}$

${\left(\frac{a+ib}{a-ib}\right)}^2={\left\{\frac{{(a+ib)}^2}{a^2+b^2}\right\}}^2$ [multiplying dividing by complex conjugate of $a-ib$]

${\left(\frac{a-ib}{a+ib}\right)}^2={\left\{\frac{{(a-ib)}^2}{a^2+b^2}\right\}}^2$ [multiplying dividing by complex conjugate of $a+ib$]

subtracting the two

$=\frac{{(a+ib)}^4-{(a-ib)}^4}{{(a^2+b^2)}^2}$

$=\frac{a^4{+}^4{C}_1{a}^3\left(ib\right){+}^4{C}_2{a}^2{\left(ib\right)}^2{+}^4{C}_3{a}^{}{\left(ib\right)}^3{+}^4{C}_4{\left(ib\right)}^4}{(a^2+b^2{)}^2}$ $\frac{-\left\{a^4{-}^4{C}_1{a}^3\left(ib\right){+}^4{C}_2{a}^2{\left(ib\right)}^2{+}^4{C}_{3}a{\left(ib\right)}^3{+}^4{C}_4{\left(ib\right)}^4\right\}}{{(a^2+b^2)}^2}$

$=\frac{8a^3\left(ib\right)+8a{\left(ib\right)}^3}{{(a^2+b^2)}^2}=\frac{8a^{3}bi-8a{b}^{3}i}{{(a^2+b^2)}^2}$

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

$Re\left(z\right)=0Im\left(z\right)=\frac{8ab(a^2-b^2)}{{(a^2+b^2)}^2}$