Question

If $(x+iy)(p+iq)=(x^2+y^2)i$, then

• A. $p=x,q=y$
• B. $p=x^2,q=y^2$
• C. $x=q,y=p$
• D. none of these

Answer 1

The correct option is C

$x=q,y=p$

Explanation for the correct answer:

To find the value of $p,q$ :

Given,

$\begin{array}{rcl}(x+iy)(p+iq)& =& (x^2+y^2)i\\ & \Rightarrow & xp+pyi+xqi-yq=\left(x^2+y^2\right)i\\ & \Rightarrow & xp-yq+(py+xq)i=(x^2+y^2)i\end{array}$

Now, on comparing the real and imaginary part

$\begin{array}{rcl}xp-yq& =& 0\\ & \Rightarrow & xp=yq\\ & \Rightarrow & x=\frac{yq}{p}\\ (py+xq)& =& (x^2+y^2)\end{array}$

Substitute $\begin{array}{rcl}x& =& \frac{yq}{p}\end{array}$

$\begin{array}{rcl}(py+\frac{y{q}^2}{p})& =& {\left(\frac{yq}{p}\right)}^2+y^2\Rightarrow \frac{y(p^2+q^2)}{p}=\frac{y^2(q^2+p^2)}{p^2}\\ & \Rightarrow & \frac{y}{p}=1\\ & \Rightarrow & y=p\end{array}$

Substitute $y$ in $\begin{array}{rcl}x& =& \frac{yq}{p}\end{array}$

Then $x=q$

Hence, the correct option is C.