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

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4-1.Complex numbers

easy

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

$p = x,q = y$

$p = {x^2},\,\,q = {y^2}$

$x = q,y = p$

None of these

Solution

(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 $x = \lambda q,y = \lambda p$
$xq + yp = {x^2} + {y^2}$

==> $\lambda = {\lambda ^2}$==> $\lambda = 1$

$x = q,y = p$.

Standard 11

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(IIT-1976)

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

Solution

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Find real $\theta$ such that

$\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is purely real.

$\frac{{3 + 2i\sin \theta }}{{1 – 2i\sin \theta }}$will be real, if $\theta $ =

[Where $n$ is an integer]