If complex numbers (x - 2y) + i(3x - y) and ( | vedclass

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

Question

(B) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.Given $(x - 2y) + i(3x - y) = \overline{(2x - y) +

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(B) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.Given $(x - 2y) + i(3x - y) = \overline{(2x - y) + i(x - y + 6)}$.This implies $(x - 2y) + i(3x - y) = (2x - y) - i(x - y + 6)$.Comparing the real and imaginary parts:Real part: $x - 2y = 2x - y \Rightarrow x + y = 0$.Imaginary part: $3x - y = -(x - y + 6)$ $\Rightarrow 3x - y = -x + y - 6$ $\Rightarrow 4x - 2y = -6$ $\Rightarrow 2x - y = -3$.Adding the two equations: $(x + y) + (2x - y) = 0 - 3$ $\Rightarrow 3x = -3$ $\Rightarrow x = -1$.Substituting $x = -1$ into $x + y = 0$,we get $y = 1$.Thus,$|x + iy| = |-1 + i| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

If complex numbers $(x - 2y) + i(3x - y)$ and $(2x - y) + i(x - y + 6)$ are conjugates of each other,then $|x + iy|$ is $(x, y \in \mathbb{R})$.

Similar Questions

If ${z_1}, {z_2} \in \mathbb{C}$,then which of the following is true?

The value of $|z - 5|$ if $z = x + iy$ is:

$A$ real value of $x$ will satisfy the equation $\left(\frac{3-4ix}{3+4ix}\right) = \alpha - i\beta$ (where $\alpha, \beta$ are real),if

If $x + iy = \sqrt{\frac{a + ib}{c + id}}$,then $(x^2 + y^2)^2 = $

Let $\alpha$ be a fixed non-zero complex number with $|\alpha| < 1$ and $w = \frac{z-\alpha}{1-\bar{\alpha}z}$,where $z$ is a complex number. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module