Mathematics · 2021
JEE Main · 27 July 2021, Shift 1 · Q67
Let C be the set of all complex numbers. Let S_1={z ∈ C | | z-3- 2 i|^2=8}, S_2={z ∈ C ∣ Re (z) ≥ 5} and S_3={z ∈ C | | z-z̄ ∣ ≥ 8}. Then the number…
Let $\displaystyle \mathbf{C}$ be the set of all complex numbers. Let
$$\begin{aligned}
& \mathrm{S}_1=\left\{z \in \mathrm{C}| | z-3-\left.2 i\right|^2=8\right\}, \\
& \mathrm{S}_2=\{z \in \mathrm{C} \mid \operatorname{Re}(z) \geqslant 5\} \text { and } \\
& \mathrm{S}_3=\{z \in \mathrm{C}| | z-\bar{z} \mid \geqslant 8\} .
\end{aligned}
$$
Then the number of elements in $\displaystyle \mathrm{S}_1 \cap \mathrm{~S}_2 \cap \mathrm{~S}_3$ is equal to :
Official answer
(2)
$\displaystyle 1$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.