Published by:
CGP EDU Academic Team
Published on: August 14, 2026
Let n denote the number of solutions of the equation 
, where z is a complexnumber. Then the value of 
is equal to
A
1
B
C
D
2
Text Solution
Verified by ExpertsThe correct answer is:
B
Put z = x + iy
x 2 − y 2 + 2ixy + 3(x − iy) = 0
(x 2 − y 2 + 3x) + i(2xy − 3y) = 0 + i0
x 2 – y = + 3x = 0……(1)
2xy − 3y = 0 ….(2)
Put 
in equation (1)
Put 
Put y = 0 
x 2 − 0 + 3x = 0
x = 0, −3
(x, y) = (0, 0), (−3, 0)
No of solutions = n = 4
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 the lines and (here ) be normal to a circle . If the line is tangent to this circle , the…
Let a complex number ,satisfy Then, the largest value of is equal to
The least value of where is complex number which satisfies the inequality , is equal to:
The area of the triangle with vertices and is:
Let and be three sets defined as
Then the set
If the equation represents a circle where are real constants then which of the following conditio…