A real value of x will satisfy the equation l | vedclass

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

Question

(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}\right)$.Taking the modulus on both sides,we get:$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}\

Vedclass · Accepted Answer

$\alpha^2 + \beta^2 = 1$

Vedclass · Answer

(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}ight)$.Taking the modulus on both sides,we get:$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}ight|$.Since the modulus of a quotient is the quotient of the moduli,we have:$|\alpha - i\beta| = \frac{|3-4ix|}{|3+4ix|}$.We know that for any complex number $z = a + ib$,$|z| = \sqrt{a^2 + b^2}$.Thus,$\sqrt{\alpha^2 + (-\beta)^2} = \frac{\sqrt{3^2 + (-4x)^2}}{\sqrt{3^2 + (4x)^2}}$.$\sqrt{\alpha^2 + \beta^2} = \frac{\sqrt{9 + 16x^2}}{\sqrt{9 + 16x^2}}$.Since $x$ is a real value,$9 + 16x^2 
eq 0$,so the ratio is $1$.$\sqrt{\alpha^2 + \beta^2} = 1$.Squaring both sides,we get $\alpha^2 + \beta^2 = 1$.

Column-$I$	Column-$II$
$A. Z_1 Z_2$	$1. \text{imaginary axis}$
$B. Z_1 + Z_2 = 0$	$2. \text{Im}(-Z_2)$
$C. \text{Im}(Z_1)$	$3. \|Z_1\|^2$
$D. \text{Re}(Z_1)$	$4. \text{Re}(Z_2)$

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

Similar Questions

$\left| {(1 + i)\frac{{(2 + i)}}{{(3 + i)}}} \right| = $

If $Z_1$ and $Z_2$ are conjugate complex numbers, match the items in Column-$I$ with Column-$II$:
Column-$I$ Column-$II$
$A. Z_1 Z_2$ $1. \text{imaginary axis}$
$B. Z_1 + Z_2 = 0$ $2. \text{Im}(-Z_2)$
$C. \text{Im}(Z_1)$ $3. |Z_1|^2$
$D. \text{Re}(Z_1)$ $4. \text{Re}(Z_2)$

If $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$,then $a^2 + b^2$ is

If $\left|z+\frac{2}{z}\right|=2$,then the maximum value of $|z|$ is

If $x + iy = \frac{3}{2 + \cos \theta + i\sin \theta}$,then ${x^2} + {y^2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module