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Complex Numbers Flashcards

6 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. What is the imaginary part of the complex number (2 + 3i)² / (1 + 2i)?

    Answer: 22/5

    First compute (2 + 3i)² = 4 + 12i + 9i² = 4 + 12i − 9 = −5 + 12i. Then divide by (1 + 2i) by multiplying numerator and denominator by the conjugate (1 − 2i): (−5 + 12i)(1 − 2i) / (1 + 4) = (−5 + 10i + 12i − 24i²) / 5 = (−5 + 22i + 24) / 5 = (19 + 22i) / 5. The imaginary part is 22/5.

  2. What is the value of i^(−15)?

    Answer: i

    First find i^15: since powers of i cycle with period 4, 15 mod 4 = 3, so i^15 = i³ = −i. Therefore i^(−15) = 1/(−i). Multiply numerator and denominator by i: i / (−i²) = i / (−(−1)) = i / 1 = i.

  3. If a and b are real numbers such that (a + bi)² = −8 + 6i, what is |a + b|?

    Answer: 4

    Expand: (a + bi)² = a² − b² + 2abi = −8 + 6i. So a² − b² = −8 and 2ab = 6, meaning ab = 3. Then (a² + b²)² = (a² − b²)² + (2ab)² = 64 + 36 = 100, so a² + b² = 10. Solving: a² = (10 − 8)/2 = 1 and b² = (10 + 8)/2 = 9. Since ab = 3 > 0, both have the same sign: (a, b) = (1, 3) or (−1, −3). In either case, |a + b| = |4| = 4.

  4. What is the sum of the imaginary parts of all complex solutions to z³ = 8i?

    Answer: 0

    Write 8i = 8e^(iπ/2). The three cube roots have modulus 2 and arguments π/6, π/6 + 2π/3 = 5π/6, and π/6 + 4π/3 = 3π/2. Their imaginary parts are: 2 sin(π/6) = 1, 2 sin(5π/6) = 1, and 2 sin(3π/2) = −2. Sum = 1 + 1 + (−2) = 0. By symmetry, the imaginary parts of the three equally-spaced cube roots always sum to zero.

  5. Let z be a complex number satisfying z + z̄ = 6 and z · z̄ = 25, where z̄ is the complex conjugate of z. What is the value of z² + z̄²?

    Answer: −14

    Use the algebraic identity z² + z̄² = (z + z̄)² − 2(z · z̄). Substituting the given values: (6)² − 2(25) = 36 − 50 = −14. Note that z · z̄ = |z|² = 25 > (z + z̄)²/4 = 9, confirming z is non-real, so z² + z̄² is indeed negative.

  6. The polynomial p(x) = x⁴ + ax² + b has real coefficients, and one of its roots is z = 1 + 2i. What is the value of a + b?

    Answer: 31

    Since p(x) has real coefficients and only even-degree terms, if 1+2i is a root then so are 1−2i, −(1+2i), and −(1−2i). The factor from 1+2i and 1−2i is x²−2x+5; the factor from −(1+2i) and −(1−2i) is x²+2x+5. Multiplying: (x²−2x+5)(x²+2x+5) = (x²+5)²−(2x)² = x⁴+10x²+25−4x² = x⁴+6x²+25. So a = 6, b = 25, and a + b = 31.