If the set left{operatorname{Re}left(frac{z-b | vedclass

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

Question

Let $z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$

Let $z=3+i y$

$\bar{z}=3-i y$

$z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i

Vedclass · Accepted Answer

$30$

Vedclass · Answer

Let $z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$

Let $z=3+i y$

$\bar{z}=3-i y$

$z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)}$

$=\frac{9+y^2+i(2 y)}{8-8 i y}$

$=\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)}$

$\operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)}$

$=\frac{9-y^2}{8\left(1+y^2\right)}$

$=\frac{1}{8}\left[\frac{10-\left(1+ y ^2\right)}{\left(1+ y ^2\right)}\right]$

$=\frac{1}{8}\left[\frac{10}{1+ y ^2}-1\right]$

$1+ y ^2 \in[1, \infty]$

$\frac{1}{1+ y ^2} \in(0,1]$

$\frac{10}{1+ y ^2} \in(0,10]$

$\frac{10}{1+ y ^2}-1 \in(-1,9]$

$\operatorname{Re}\left( z _1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right]$

$\alpha=\frac{-1}{8}, \beta=\frac{9}{8}$

$24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30$

If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C , \operatorname{Re}(z)=3\right\}$ is equal to the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to

Similar Questions

For any two complex numbers ${z_1}$and${z_2}$ and any real numbers $a$ and $b$; $|(a{z_1} - b{z_2}){|^2} + |(b{z_1} + a{z_2}){|^2} = $

Let $z$ satisfy $\left| z \right| = 1$ and $z = 1 - \vec z$.

Statement $1$ : $z$ is a real number

Statement $2$ : Principal argument of $z$ is $\frac{\pi }{3}$

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$then arg $({z_1}) - $arg $({z_2})$ is equal to

The values of $z$for which $|z + i|\, = \,|z - i|$ are

The amplitude of the complex number $z = \sin \alpha + i(1 - \cos \alpha )$ is