JEE Mains · Maths · STD 11 - 4.1 complex nubers
The area of the polygon, whose vertices are the non-real roots of the equation \(z=i z^{2}\) is
- A \(\frac{3}{2}\)
- B \(\frac{3 \sqrt{3}}{2}\)
- C \(\frac{3 \sqrt{3}}{4}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \sqrt{3}}{4}\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow\) Let \(z=x+i y, x, y \in R\) Now \(\bar{z}=i z^{2}\) then \(x-i y=i\left(x^{2}-y^{2}+2 x y i\right)\) \(x-i y=i\left(x^{2}-y^{2}\right)-2 x y\) \(\Rightarrow x=-2 x y\;and \;-y=x^{2}-y^{2}\) \(\Rightarrow x (1+2 y )=0\) \(x=0\) or \(y=-\frac{1}{2}\) Put \(x=0\) in…
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