alleryStart learning →Solve z² + 1 = 0:
- z = ±1
- z = -1
- z = ±i
- z = 1
Show answer and solution
Correct answer: C. z = ±i
We have z² = -1 → z.
So z² = ±i.
*Note: complex numbers were invented to give meaning to √(-1) = ±i.*
Indian Navy Pilot Complex Numbers and Quadratic Equations — practice questions
Explore 6 free Complex Numbers and Quadratic Equations questions in Indian Navy Pilot Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
Practise Complex Numbers and Quadratic Equations with Quill →Complex Numbers and Quadratic Equations questions and exercises with answers
Showing questions 1–6 of 6.
Solve z² + 1 = 0:Read the answer and worked solution →Compute (1 + i)⁸:Read the answer and worked solution →If z and z̄ are roots of a quadratic with real coefficients, then sum z + z̄ equals:Read the answer and worked solution →In z = a + ib, Re(z) and Im(z) areRead the answer and worked solution →The modulus $|3 + 4i|$ equals:Read the answer and worked solution →Which complex number has the larger modulus?Read the answer and worked solution →
Worked practice questions
Compute (1 + i)⁸:
- -16i
- 16
- 16i
- -16
Show answer and solution
Correct answer: B. 16
(1+i) = √2 cis(π/4).
**So (1+i)⁸ = (√2)⁸ × cis(8π/4) = 16 × cis(2π) = 16 × 1 = 16.**
If z and z̄ are roots of a quadratic with real coefficients, then sum z + z̄ equals:
- 2 Re(z) (twice real part)
- 0
- |z|
- Imaginary
Show answer and solution
Correct answer: A. 2 Re(z) (twice real part)
We have z + z̄ = (a + ib) + (a - ib).
That is z + z̄ = 2a.
So z + z̄ = 2 Re(z).
Complex conjugate pairs are roots of real-coefficient polynomials.
Sum and product of conjugates are both real.
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