Let C be the set of all complex numbers.

Question

Let $\mathbb C$ be the set of all complex numbers. Let $S_1=\{z\in\mathbb C:|z-3-2i|^2=8\}$, $S_2=\{z\in\mathbb C:\operatorname{Re}(z)\ge5\}$ and $S_3=\{z\in\mathbb C:|z-\bar z|\ge8\}$. Then the number of elements in $S_1\cap S_2\cap S_3$ is equal to

Accepted Answer

${S_1}:|z - 3 - 2i{|^2} = 8$

$$|z - 3 - 2i| = 2\sqrt 2 $$

$${(x - 3)^2} + {(y - 2)^2} = {(2\sqrt 2 )^2}$$

$${S_2}:x \ge 5$$

$${S_3}:|z - \overline z, \ge 8$$

$$|2iy| \ge 8$$

$$2|y| \ge 8$$

$$\therefore$$ $$y \ge 4$$, $$y \le  - 4$$

$$n\left( {{S_1} \cap {S_2} \cap {S_3}} \right) = 1$$

Solutions · 1