For a , let A = { z : R e ( a + z ) I m ( a + z ) } and B = { z : R e ( a + z ) I m ( a + z ) } . Then among the two statements: (S1) : If R e ( a ) , I m ( a ) 0 , then the set A contains all the real numbers. (S2) : If R e ( a ) , I m

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Solution

Q.
For $a \in ℂ$ , let $A = {z \in ℂ : R e (a + \bar{z}) > I m (\bar{a} + z)} and B = {z \in ℂ : R e (a + \bar{z}) < I m (\bar{a} + z)} .$ Then among the two statements:

(S1) : If $R e (a), I m (a) > 0$ , then the set A contains all the real numbers.

(S2) : If $R e (a), I m (a) < 0$ , then the set B contains all the real numbers. [2023]

1 only (S2) is true

2 both are true

3 only (S1) is true

4 both are false

Ans.
(4)

$Let a = x_{1} + i y_{1} and z = x + i y$

$Now Re (a + \bar{z}) > Im (\bar{a} + z)$

$\Rightarrow Re (x_{1} + i y_{1} + x - i y) > Im (x_{1} - i y_{1} + x + i y)$

$\Rightarrow Re (x_{1} + x + i (y_{1} - y)) > Im (x_{1} + x + i (y - y_{1}))$

$∴ x_{1} + x > - y_{1} + y$

$Let x_{1} = 2, y_{1} = 10, x = - 12, y = 0$

Given inequality is not valid for these values.

$S 1 is false.$

$Now Re (a + \bar{z}) < Im (\bar{a} + z)$

$\Rightarrow Re (x_{1} + x + i (y_{1} - y)) < Im (x_{1} + x + i (y - y_{1}))$

$\Rightarrow x_{1} + x < - y_{1} + y$

$Let x_{1} = - 2, y_{1} = - 10, x = 12, y = 0$

Given inequality is not valid for these values.

$S 2 is false.$

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