If alpha and beta are different complex numbe | vedclass

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

Question

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given that $|\beta|=1$,we have $|\beta|^2 = \beta \bar{\beta} = 1$.Consider the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$.Substituting $1 = \beta \bar{\beta}$ in the denominator,we get:$\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| = \left|\frac{\beta-\alpha}{\beta(\bar{\beta}-\bar{\alpha})}\right|$.Using the property of modulus $|z_1/z_2| = |z_1|/|z_2|$ and $|z_1 z_2| = |z_1||z_2|$,we have:$= \frac{|\beta-\alpha|}{ |\beta| |\bar{\beta}-\bar{\alpha}| }$.Since $|\beta|=1$ and $|\bar{z}| = |z|$,we know $|\bar{\beta}-\bar{\alpha}| = |\overline{\beta-\alpha}| = |\beta-\alpha|$.Therefore,the expression becomes $\frac{|\beta-\alpha|}{1 \cdot |\beta-\alpha|} = 1$.

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$,then $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ is equal to

Similar Questions

If $z = \frac{1}{1 - \cos \theta + i \sin \theta}$ and $\theta$ is acute,then the modulus and amplitude of $z$ respectively are:

If the point $z=(1+i)(1+2i)(1+3i) \ldots (1+10i)$ lies on a circle with centre at the origin and radius $r$,then $r^2$ is equal to

If $a > 0$ and $z = \frac{(1+i)^2}{a-i}$,where $i = \sqrt{-1}$,has a magnitude of $\frac{2}{\sqrt{5}}$,then $\bar{z}$ is

If $\alpha_1, \alpha_2, \alpha_3$ respectively denote the moduli of the complex numbers $-i, \frac{1}{3}(1+i)$ and $-1+i$,then their increasing order is

The real part of $(1 - \cos \theta + 2i\sin \theta )^{-1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module