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

(C) Let $z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}$.We calculate $|z|^2 = z \cdot \overline{z} = \left( \frac{\beta - \alpha}{1 - \overlin

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}$.We calculate $|z|^2 = z \cdot \overline{z} = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta} - \overline{\alpha}}{1 - \alpha\overline{\beta}} \right)$.Expanding the numerator: $(\beta - \alpha)(\overline{\beta} - \overline{\alpha}) = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha} = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.Expanding the denominator: $(1 - \overline{\alpha}\beta)(1 - \alpha\overline{\beta}) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \alpha\overline{\alpha}\beta\overline{\beta} = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2|\beta|^2$.Since $|\beta| = 1$,we have $|\beta|^2 = 1$.Substituting $|\beta|^2 = 1$ into the expressions:Numerator: $1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$.Denominator: $1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2(1) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2$.Since the numerator and denominator are equal,$|z|^2 = 1$,which implies $|z| = 1$.

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

Similar Questions

Let $z$ be a complex number such that $|z|+z=2+i$,where $i=\sqrt{-1}$. Then $|z|$ is equal to:

If $x-iy = \sqrt{\frac{a-ib}{c-id}}$,prove that $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.

The number of integer solutions of the equation $|1-i|^x=2^x$ is

If $x+iy = \frac{(3+2i)(4-7i)(12+13i)}{(13-12i)(2-3i)(11+3i)}$,then $x^2+y^2=$

If $|z_1| = |z_2| = \dots = |z_n| = 1$,then the value of $|z_1 + z_2 + z_3 + \dots + z_n|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module