Exams Joint Entrance Examination Mathematics Complex Numbers Question 438 Question 438 of 455 Medium MCQ
If the set { Re ( z − z ˉ + z z ˉ 2 − 3 z + 5 z ˉ ) : z ∈ C , Re ( z ) = 3 } \left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in \mathbb{C}, \operatorname{Re}(z)=3\right\} { Re ( 2 − 3 z + 5 z ˉ z − z ˉ + z z ˉ ) : z ∈ C , Re ( z ) = 3 } is equal to
the interval ( α , β ] (\alpha, \beta] ( α , β ] , then 24 ( β − α ) 24(\beta-\alpha) 24 ( β − α ) is equal to :
Solutions · 1 Post your solution Official solution
Let z 1 = ( z − z ˉ + z z ˉ 2 − 3 z + 5 z ˉ ) z_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right) z 1 = ( 2 − 3 z + 5 z ˉ z − z ˉ + z z ˉ )
Let z = 3 + i y \mathrm{z}=3+\mathrm{iy} z = 3 + iy
z ˉ = 3 − i y \bar{z}=3-i y z ˉ = 3 − i y
\[
z 1 = 2 i y + ( 9 + y 2 ) 2 − 3 ( 3 + i y ) + 5 ( 3 − i y ) = 9 + y 2 + i ( 2 y ) 8 − 8 i y = ( 9 + y 2 ) + i ( 2 y ) 8 ( 1 − i y ) Re ( z 1 ) = ( 9 + y 2 ) − 2 y 2 8 ( 1 + y 2 ) \begin{aligned}
& z_1=\frac{2 i y+\left(9+y^2\right)}{2-3(3+i y)+5(3-i y)} \\\\
& =\frac{9+y^2+i(2 y)}{8-8 i y} \\\\
& =\frac{\left(9+y^2\right)+i(2 y)}{8(1-i y)} \\\\
& \operatorname{Re}\left(z_1\right)=\frac{\left(9+y^2\right)-2 y^2}{8\left(1+y^2\right)}
\end{aligned} z 1 = 2 − 3 ( 3 + i y ) + 5 ( 3 − i y ) 2 i y + ( 9 + y 2 ) = 8 − 8 i y 9 + y 2 + i ( 2 y ) = 8 ( 1 − i y ) ( 9 + y 2 ) + i ( 2 y ) Re ( z 1 ) = 8 ( 1 + y 2 ) ( 9 + y 2 ) − 2 y 2
\]
\[
= 9 − y 2 8 ( 1 + y 2 ) = 1 8 [ 10 − ( 1 + y 2 ) ( 1 + y 2 ) ] = 1 8 [ 10 1 + y 2 − 1 ] 1 + y 2 ∈ [ 1 , ∞ ] 1 1 + y 2 ∈ ( 0 , 1 ] 10 1 + y 2 ∈ ( 0 , 10 ] 10 1 + y 2 − 1 ∈ ( − 1 , 9 ] Re ( z 1 ) ∈ ( − 1 8 , 9 8 ] α = − 1 8 , β = 9 8 24 ( β − α ) = 24 ( 9 8 + 1 8 ) = 30 \begin{aligned}
& =\frac{9-y^2}{8\left(1+y^2\right)} \\\\
& =\frac{1}{8}\left[\frac{10-\left(1+y^2\right)}{\left(1+y^2\right)}\right] \\\\
& =\frac{1}{8}\left[\frac{10}{1+y^2}-1\right] \\\\
& 1+y^2 \in[1, \infty] \\\\
& \frac{1}{1+y^2} \in(0,1] \\\\
& \frac{10}{1+y^2} \in(0,10] \\\\
& \frac{10}{1+y^2}-1 \in(-1,9] \\\\
& \operatorname{Re}\left(\mathrm{z}_1\right) \in\left(\frac{-1}{8}, \frac{9}{8}\right] \\\\
& \alpha=\frac{-1}{8}, \beta=\frac{9}{8} \\\\
& 24(\beta-\alpha)=24\left(\frac{9}{8}+\frac{1}{8}\right)=30
\end{aligned} = 8 ( 1 + y 2 ) 9 − y 2 = 8 1 [ ( 1 + y 2 ) 10 − ( 1 + y 2 ) ] = 8 1 [ 1 + y 2 10 − 1 ] 1 + y 2 ∈ [ 1 , ∞ ] 1 + y 2 1 ∈ ( 0 , 1 ] 1 + y 2 10 ∈ ( 0 , 10 ] 1 + y 2 10 − 1 ∈ ( − 1 , 9 ] Re ( z 1 ) ∈ ( 8 − 1 , 8 9 ] α = 8 − 1 , β = 8 9 24 ( β − α ) = 24 ( 8 9 + 8 1 ) = 30
\] Answer the question to unlock