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UPCET / UPSEE Complex Numbers and Quadratic Equations — practice questions — page 2

Explore 49 free Complex Numbers and Quadratic Equations questions in UPCET / UPSEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Complex Numbers and Quadratic Equations questions and exercises with answers

Showing questions 31–49 of 49.

Find principal argument of -1 - i:Read the answer and worked solution →For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) = 0?Read the answer and worked solution →Solve: z² + z + 1 = 0:Read the answer and worked solution →If z1, z2, z3 are vertices of an equilateral triangle, then:Read the answer and worked solution →Find (3 + 4i)/(1 + 2i):Read the answer and worked solution →The value of i⁴ⁿ⁺³ (n a positive integer) isRead the answer and worked solution →The modulus of the complex number 3 + 4i isRead the answer and worked solution →The conjugate of (2 − 3i) isRead the answer and worked solution →If z = 1 + i, then z² equalsRead the answer and worked solution →The argument (principal value) of the complex number i isRead the answer and worked solution →The multiplicative inverse of (3 + 4i) isRead the answer and worked solution →For complex numbers, which relation is NOT defined?Read the answer and worked solution →The roots of x² + 1 = 0 areRead the answer and worked solution →If z = a + ib is purely imaginary, thenRead the answer and worked solution →The value of (1 + i)/(1 − i) isRead the answer and worked solution →|z₁ z₂| equalsRead the answer and worked solution →In z = a + ib, Re(z) and Im(z) areRead the answer and worked solution →Real or imaginary?Read the answer and worked solution →Which one does NOT belong?Read the answer and worked solution →

Worked practice questions

Find principal argument of -1 - i:

  1. -π/4
  2. -3π/4
  3. 3π/4
  4. π/4
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Correct answer: B. -3π/4

-1 - i is at (-1, -1), in third quadrant. Principal argument range (-π, π]: arg = -3π/4 (or equivalently +5π/4, but principal value is in (-π, π]).

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For roots of unity ω^n = 1: which is the SMALLEST n such that 1 + ω + ω² + ... + ω^(n-1) = 0?

  1. 2
  2. Only n = 3
  3. 1
  4. Any n > 1
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Correct answer: D. Any n > 1

Sum 1 + ω + ω² + ... + ω^(n-1) = (ω^n - 1)/(ω - 1) = 0 when ω is a primitive n-th root (ω ≠ 1). **So for any n ≥ 2 (with non-trivial primitive root), sum is 0.**

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Solve: z² + z + 1 = 0:

  1. z = ±i
  2. z = ±1
  3. z = (-1 ± i√3)/2 (complex cube roots of unity other than 1)
  4. No solution
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Correct answer: C. z = (-1 ± i√3)/2 (complex cube roots of unity other than 1)

z = (-1 ± √(1 - 4))/2 = (-1 ± √(-3))/2 = (-1 ± i√3)/2. These are ω and ω² (complex cube roots of unity).

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