Question

if $(a+ib)(x+iy)=(a^2+b^2)i,a^2+b^2\ne 0$, then show that $x=b$ and $y=a$

Answer 1

$(a+ib)(x+iy)=(a^2+b^2)i$

$ax-by+ibx+iay=(a^2+b^2)i$

$ax-by+i(bx+ay)=(a^2+b^2)i$

By comparing both sides

$ax-by=0$ & $bx+ay=a^2+b^2$

$ax=by$

$\frac{x}{y}=\frac{b}{a}$

$bx=\frac{b^{2}y}{a}$ ______ (1)

$bx+ay=a^2+b^2$

using Result (1)

$\frac{b^{2}y}{a}+ay=a^2+b^2$

$y\left(\frac{b^2}{a}\right)=a^2+b^2$

$y\frac{(b^2+a^2)}{a}=a^2+b^2$

$\frac{y}{a}1$

$y=a$ and from (1) $x=b$