Let S = { z = x + i y : 2 z - 3 i 4 z + 2 i is a real number } . Then which of the following is NOT correct? [2023]

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Q.
Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct [2023]

1 $(x, y) = (0, - \frac{1}{2})$

2 $y + x^{2} + y^{2} \neq - \frac{1}{4}$

3 $y \in (- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

4 $x = 0$

Ans.
(1)

$As, \frac{2 z - 3 i}{4 z + 2 i} is a real number.$

$∴$ $\frac{2 z - 3 i}{4 z + 2 i} = \frac{\bar{2 z - 3 i}}{4 z + 2 i} \Rightarrow \frac{2 z - 3 i}{4 z + 2 i} = \frac{2 \bar{z} + 3 i}{4 \bar{z} - 2 i}$

$\Rightarrow (2 z - 3 i) (4 \bar{z} - 2 i) = (2 \bar{z} + 3 i) (4 z + 2 i)$

$\Rightarrow 8 z \bar{z} - 4 i z - 12 i \bar{z} - 6 = 8 z \bar{z} + 4 i \bar{z} + 12 i z - 6$

$\Rightarrow - 4 i (z + \bar{z}) - 12 i (z + \bar{z}) = 0$

$\Rightarrow 16 i (z + \bar{z}) = 0 \Rightarrow z + \bar{z} = 0 \Rightarrow 2 x = 0 \Rightarrow x = 0$

$As, y can not be \frac{- 1}{2} .$

$∴$ $Option (1) is not possible.$

Solution

Q.
Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct [2023]

1 $(x, y) = (0, - \frac{1}{2})$

2 $y + x^{2} + y^{2} \neq - \frac{1}{4}$

3 $y \in (- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

4 $x = 0$

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Solution

Q. Let S={z=x+iy:2z-3i4z+2i is a real number}. Then which of the following is NOT correct [2023]

1 (x,y)=(0,-12)

2 y+x2+y2≠-14

3 y∈(-∞,-12)∪(-12,∞)

4 x=0

Ans. (1) As, 2z-3i4z+2i is a real number. ∴ 2z-3i4z+2i=2z-3i¯4z+2i⇒2z-3i4z+2i=2z¯+3i4z¯-2i ⇒(2z-3i)(4z¯-2i)=(2z¯+3i)(4z+2i) ⇒8zz¯-4iz-12iz¯-6=8zz¯+4iz¯+12iz-6 ⇒-4i(z+z¯)-12i(z+z¯)=0 ⇒16i(z+z¯)=0⇒z+z¯=0⇒2x=0⇒x=0 As, y can not be -12. ∴ Option (1) is not possible.

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Q.
Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number} .$ Then which of the following is NOT correct [2023]

1 $(x, y) = (0, - \frac{1}{2})$

2 $y + x^{2} + y^{2} \neq - \frac{1}{4}$

3 $y \in (- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

4 $x = 0$