यदि समुच्चय left{operatorname{Re}left(frac{ma | vedclass

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

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