MCQ
If $(x + iy)(p + iq) = ({x^2} + {y^2})i$, then
- A$p = x,q = y$
- B$p = {x^2},\,\,q = {y^2}$
- ✓$x = q,y = p$
- DNone of these
✓
Answer
Correct option: C.
$x = q,y = p$
c
(c) $(x + iy)(p + iq) = ({x^2} + {y^2})i$
==> $(xp - yq) + i(xq + yp) = ({x^2} + {y^2})i$
==> $xp - yq = 0,xq + yp = {x^2} + {y^2}$
==> $\frac{x}{q} = \frac{y}{p}$and$xq + yp = {x^2} + {y^2}$
Let $\frac{x}{q} = \frac{y}{p} = \lambda $. then $x = \lambda q,y = \lambda p$
$xq + yp = {x^2} + {y^2}$
(c) $(x + iy)(p + iq) = ({x^2} + {y^2})i$
==> $(xp - yq) + i(xq + yp) = ({x^2} + {y^2})i$
==> $xp - yq = 0,xq + yp = {x^2} + {y^2}$
==> $\frac{x}{q} = \frac{y}{p}$and$xq + yp = {x^2} + {y^2}$
Let $\frac{x}{q} = \frac{y}{p} = \lambda $. then $x = \lambda q,y = \lambda p$
$xq + yp = {x^2} + {y^2}$
==> $\lambda = {\lambda ^2}$==> $\lambda = 1$
$x = q,y = p$.
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