Question

If $\frac{(a+i{)}^2}{2a-i}=p+iq,$ where $a\in R,$ then the value of $p^2+q^2$ is

• A. $\frac{(a^2-1{)}^2}{4a^2+1}$
• B. $\frac{(a^2+1{)}^2}{4a^2+1}$
• C. $\frac{(a^2-1{)}^2}{2a^2+1}$
• D. $\frac{(a^2+1{)}^2}{2a^2+1}$

Answer 1

The correct option is B $\frac{(a^2+1{)}^2}{4a^2+1}$

Given : $\frac{(a+i{)}^2}{2a-i}=p+iq$
Taking conjugate on both sides, we get
$\overline{p+iq}=\overline{\left(\frac{(a+i{)}^2}{(2a-i)}\right)}\Rightarrow p-iq=\frac{\overline{(a+i{)}^2}}{\overline{(2a-i)}}(\because \overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}})\Rightarrow p-iq=\frac{(a-i{)}^2}{(2a+i)}(\because \overline{\left(z^n\right)}=(\overline{z}{)}^n)$

Now, $p^2+q^2$
$=(p+iq)\times (p-iq)=\frac{(a+i{)}^2}{2a-i}\times \frac{(a-i{)}^2}{(2a+i)}=\frac{{(a^2+1)}^2}{4a^2+1}$