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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

z

is a complex number satisfying

z^4 + z^3 + 2z^2 + z + 1 = 0

, then the set of possible values of

|z|

\{1, 2\}

\{1\}

correct

\{1, 2, 3\}

\{1, 2, 3, 4\}

Solution

Step 1: Factorise the quartic. Group the terms so a common quadratic factor appears:

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)

= (z^2 + z + 1)(z^2 + 1)

Step 2: Solve

z^2 + z + 1 = 0

using the quadratic formula:

z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

These are the non-real cube roots of unity

\omega

and

\omega^2

. Their modulus:

|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

z^2 + 1 = 0

z^2 = -1 \implies z = \pm i, \qquad |z| = 1

Step 4: All four roots have modulus

1

, so the set of possible values of

|z|

\{1\}

. Correct answer: (2)

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