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Modulus of Roots z⁴+z³+2z²+z+1=0 | Complex Numbers JEE

JEE Maths question with a full step-by-step solution.

Question
If zz is a complex number satisfying z4+z3+2z2+z+1=0z^4 + z^3 + 2z^2 + z + 1 = 0, then the set of possible values of ∣z∣|z| is
A{1,2}\{1, 2\}
B{1}\{1\}correct
C{1,2,3}\{1, 2, 3\}
D{1,2,3,4}\{1, 2, 3, 4\}
Solution
Step 1: Factorise the quartic. Group the terms so a common quadratic factor appears:
z4+z3+2z2+z+1=(z4+z3+z2)+(z2+z+1)=z2(z2+z+1)+(z2+z+1)z^4 + z^3 + 2z^2 + z + 1 = (z^4 + z^3 + z^2) + (z^2 + z + 1) = z^2(z^2 + z + 1) + (z^2 + z + 1)
=(z2+z+1)(z2+1)= (z^2 + z + 1)(z^2 + 1)
 Step 2: Solve z2+z+1=0z^2 + z + 1 = 0 using the quadratic formula:
z=−1±1−42=−1±i32z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
 These are the non-real cube roots of unity ω\omega and ω2\omega^2. Their modulus:
∣z∣=(−12)2+(32)2=14+34=1=1|z| = \sqrt{\left(-\tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt 3}{2}\right)^2} = \sqrt{\tfrac{1}{4} + \tfrac{3}{4}} = \sqrt{1} = 1
 Step 3: Solve z2+1=0z^2 + 1 = 0:
z2=−1  ⟹  z=±i,∣z∣=1z^2 = -1 \implies z = \pm i, \qquad |z| = 1
 Step 4: All four roots have modulus 11, so the set of possible values of ∣z∣|z| is {1}\{1\}. Correct answer: (2)
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