JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 4)
Let C be the set of all complex numbers. Let
$${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $$
$${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $$ and
$${S_3} = \{ z \in C||z - \overline z | \ge 8\} $$.
Then the number of elements in $${S_1} \cap {S_2} \cap {S_3}$$ is equal to :
$${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $$
$${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $$ and
$${S_3} = \{ z \in C||z - \overline z | \ge 8\} $$.
Then the number of elements in $${S_1} \cap {S_2} \cap {S_3}$$ is equal to :
1
0
2
Infinite
Explanation
$${S_1}:|z - 3 - 2i{|^2} = 8$$
$$|z - 3 - 2i| = 2\sqrt 2 $$
$${(x - 3)^2} + {(y - 2)^2} = {(2\sqrt 2 )^2}$$
$${S_2}:x \ge 5$$
$${S_3}:|z - \overline z | \ge 8$$
$$|2iy| \ge 8$$
$$2|y| \ge 8$$
$$\therefore$$ $$y \ge 4$$, $$y \le - 4$$
_27th_July_Morning_Shift_en_4_1.png)
$$n\left( {{S_1} \cap {S_2} \cap {S_3}} \right) = 1$$
$$|z - 3 - 2i| = 2\sqrt 2 $$
$${(x - 3)^2} + {(y - 2)^2} = {(2\sqrt 2 )^2}$$
$${S_2}:x \ge 5$$
$${S_3}:|z - \overline z | \ge 8$$
$$|2iy| \ge 8$$
$$2|y| \ge 8$$
$$\therefore$$ $$y \ge 4$$, $$y \le - 4$$
$$n\left( {{S_1} \cap {S_2} \cap {S_3}} \right) = 1$$
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