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Let z 1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then ?

Let z 1 and z2 be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle, then ?

Grade:11

3 Answers

Latika Leekha
askIITians Faculty 165 Points
6 years ago
Hello student,
Given that z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex.
So, we have z1 + z2 = -a, z1z2 = b (sum and product of roots)
So, z2 = z1 ( cos 60° + i sin 60°)
= z1 (1/2 + √3/2 i)
So, 2z2 – z1 = √3 i z1
This gives, (2z2 -z1)2 = -3z12
Hence, (z12 + z22) = z1z2
So, a2 – 2b = b
this gives a2 = 3b.
Anupam Verma
26 Points
3 years ago
As z1, z2 are roots of z2 + az + bz1 + z2 = –a and z1 z2 = bAgain 0, z1, z2 are vertices of an equilateral triangle.Therefore, 02 + z12 + z22 = 0z1 + z1z2 + 0z2 = 0z12 + z22 = z1z2(z1 + z2)2 = 3z1z2
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the attached solution to your problem below.
 
Given that z1 and z2 be two roots of the equation z2 + az + b = 0, z being complex.
So, we have z1 + z2 = -a, z1z2 = b (sum and product of roots)
So, z2 = z1 ( cos 60° + i sin 60°) = z1 (1/2 + √3/2 i)
So, 2z2 – z1 = √3 i z1
This gives, (2z2 -z1)2 = -3z12
Hence, (z12 + z22) = z1z2
So, a2 – 2b = b
this gives a2 = 3b
 
Hope it helps.
Thanks and regards,
Kushagra

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