Test Index

Complex Numbers Part 2

Section: Mathematics

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Question : 63 of 106

Marks: +1, -0

If the set

\{ \operatorname{Re}\left( \frac{z-\overline{z}+z\overline{z}}{2-3z+5\overline{z}} \right) : z \in C, \operatorname{Re}(z)=3 \}

is equal to the interval

(\alpha, \beta]

24(\beta - \alpha)

[15-Apr-2023 shift 1]

36
27
30
42

Solution:

Let

z_1 = \left( \frac{z - \overline{z} + z \overline{z}}{2-3z+5\overline{z}} \right)

Let

z=3+

iy

\overline{z}=3-

iy

\;z_1 = \frac{2 i y + (9+y^2)}{2-3(3+ i y)+5(3- i y)}

\; = \frac{9+y^2+ i(2y)}{8-8 i y}

\; = \frac{(9+y^2)+i(2y)}{8(1- i y)}

\operatorname{Re}(z_1) = \frac{(9+y^2)-2y^2}{8(1+y^2)}

\; = \frac{9-y^2}{8(1+y^2)}

\; = \frac{1}{8} \left[ \frac{10 - (1+y^2)}{1+y^2} \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}(z_1) \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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