Published by:
CGP EDU Academic Team
Published on: August 14, 2026
The area of the polygon, whose vertices are the non-real roots of the equation
:
Text Solution
Verified by ExpertsThe correct answer is:
A
Let z = x + iy, x, y
R
Now 
then x-iy = i(x 2 -y 2 + 2xyi)
x - iy = i (x 2 - y 2 ) - 2 xy


x = 0 or 
Put x = 0 in -y = x 2 -y 2
We get y = y 2 
y = 0,1
Similarly
Put
in 





Area 

──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
Let and . Then, :
Let a circle C in complex plane pass through the points z 1 = 3 + 4i, z 2 = 4 + 3i and z 3 = 5i. If…
Let z 1 and z 2 be two complex numbers such that and . Then
Let and . Then A B is:
The number of points of intersection of |z-(4 + 3i)| = 2 and |z| + |z-4| = 6, z C is :
Let and be the roots of the equation x 2 + (2i - 1) = 0. Then, the value of is equal to: