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

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(C) Given: $(x + iy)(p + iq) = (x^2 + y^2)i$Expanding the left side: $(xp - yq) + i(xq + yp) = 0 + i(x^2 + y^2)$Comparing real and imaginary parts:$xp

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$x = q, y = p$

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(C) Given: $(x + iy)(p + iq) = (x^2 + y^2)i$Expanding the left side: $(xp - yq) + i(xq + yp) = 0 + i(x^2 + y^2)$Comparing real and imaginary parts:$xp - yq = 0 \implies xp = yq \implies \frac{x}{q} = \frac{y}{p} = \lambda$$xq + yp = x^2 + y^2$Substituting $x = \lambda q$ and $y = \lambda p$ into the second equation:$(\lambda q)q + (\lambda p)p = (\lambda q)^2 + (\lambda p)^2$$\lambda(q^2 + p^2) = \lambda^2(q^2 + p^2)$Since $x^2 + y^2 
eq 0$ (assuming non-zero complex number),we have $\lambda = \lambda^2$,which implies $\lambda = 1$.Therefore,$\frac{x}{q} = 1 \implies x = q$ and $\frac{y}{p} = 1 \implies y = p$.

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

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