Question

If $x+iy=\sqrt{\frac{a+ib}{c+id}},$ then $x^4+y^4+2x^2{y}^2$ is equal to

• A. $\frac{(a+b{)}^2}{(c+d{)}^2}$
• B. $\frac{a^2+b^2}{c^2+d^2}$
• C. $\frac{(a-b{)}^2}{(c-d{)}^2}$
• D. $\frac{a^2-b^2}{c^2-d^2}$

Answer 1

The correct option is B $\frac{a^2+b^2}{c^2+d^2}$

We have,
$x+iy=\sqrt{\frac{a+ib}{c+id}}\Rightarrow (x+iy{)}^2=\frac{a+ib}{c+id}\Rightarrow \overline{(x+iy{)}^2}=\overline{\left(\frac{a+ib}{c+id}\right)}\Rightarrow (x-iy{)}^2=\frac{a-ib}{c-id}[\because \overline{(z{)}^n}=(\overline{z}{)}^n,\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}]$

Therefore,
$x^4+y^4+2x^2{y}^2=(x^2+y^2{)}^2=[(x+iy)(x-iy){]}^2=(x+iy{)}^{\times }(x-iy{)}^2=\frac{a+ib}{c+id}\times \frac{a-ib}{c-id}=\frac{a^2+b^2}{c^2+d^2}$