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Let $z=x+i y$ be a complex number,
$A=\{z /|z| \leq 2\} \text { and } B=\{z /(1-i) z+(1+i) \bar{z} \geq 4\}$
Then which one of the following options belongs to $A \cap B$ ?
Options:
$A=\{z /|z| \leq 2\} \text { and } B=\{z /(1-i) z+(1+i) \bar{z} \geq 4\}$
Then which one of the following options belongs to $A \cap B$ ?
Solution:
2096 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3}+\frac{1}{2} i$
We have, $|z| \leq 2$
$\begin{aligned} & \Rightarrow \quad(1-i)(x+i y)+(1+i)(x-i y) \geq 4 \\ & \Rightarrow \quad x+i y-i x+y+x-i y+i x+y \geq 4\end{aligned}$
$\begin{aligned} & \text { So, } A \cap B=\left\{x^2+y^2 \leq 4\right\} \cap\{x+y \geq 2\} \\ & \therefore \quad z=\sqrt{3}+\frac{1}{2} i\end{aligned}$
$\begin{aligned} & \Rightarrow \quad(1-i)(x+i y)+(1+i)(x-i y) \geq 4 \\ & \Rightarrow \quad x+i y-i x+y+x-i y+i x+y \geq 4\end{aligned}$
$\begin{aligned} & \text { So, } A \cap B=\left\{x^2+y^2 \leq 4\right\} \cap\{x+y \geq 2\} \\ & \therefore \quad z=\sqrt{3}+\frac{1}{2} i\end{aligned}$
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