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  • For , let <img alt="\mathrm{A}=\{\mathrm{z} \in \mathrm{C}:

For \mathrm{a} \in \mathrm{C}, let \mathrm{A}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})>\operatorname{Im}(\overline{\mathrm{a}}+\mathrm{z})\} and \mathrm{B}=\{\mathrm{z} \in \mathrm{C}: \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})<\operatorname{Im}(\overline{\mathrm{a}}+\mathrm{z})\}. The among the two statements:
(S1) : If Re (a), Im (a) > 0 , then the set A contains all the real numbers
(S2) : If Re (a), Im (a) < 0, then the set B contains all the real numbers,
 

Option: 1

only (S1) is true


Option: 2

both are false


Option: 3

only (S2) is true


Option: 4

both are true


Answers (1)

best_answer

Let \mathrm{a}=\mathrm{x}_{1}+\mathrm{iy}_{1} \mathrm{z}=\mathrm{x}+\mathrm{iy}

Now \: \operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)
\therefore \mathrm{x}_{1}+\mathrm{x}>-\mathrm{y}_{1}+\mathrm{y}
\mathrm{x}_{1}=2, \mathrm{y}_{1}=10, \mathrm{x}=-12, \mathrm{y}=0

Given inequality is not valid for these values.
S1 is false.

Now \operatorname{Re}(\mathrm{a}+\overline{\mathrm{z}})<\operatorname{Im}(\overline{\mathrm{a}}+\mathrm{z})
\mathrm{x}_{1}+\mathrm{x}<-\mathrm{y}_{1}+\mathrm{y}
\mathrm{x}_{1}=-2, \mathrm{y}_{1}=-10, \mathrm{x}=12, \mathrm{y}=0

Given inequality is not valid for these values.
S2 is false.

Posted by

shivangi.shekhar

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