4-1.Complex numbers
easy

જો $(x + iy)(p + iq) = ({x^2} + {y^2})i$ તો

A

$p = x,q = y$

B

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

C

$x = q,y = p$

D

એકપણ નહીં.

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
Mathematics

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