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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$x = q, y = p$

Vedclass · Answer

(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

Similar Questions

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$,where $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then $(a, b)$ is equal to

If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$,then $(x, y)$ is equal to

The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1=$

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$,then the values of $x$ and $y$ are:

The value of $\frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}}$,where $i=\sqrt{-1}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module