Compute 1/i (write in standard form a + bi). — −i.Multiply top and bottom by i: i/i² = i/(−1) = −i. So 1/i = −i.
Which expression is purely real (no imaginary component)? — (1+i)(1−i).(1+i)(1−i) = 1² + 1² = 2, which is purely real. The product of conjugates is always real.
If z = a + bi, what is z · z̄ (z times its conjugate)? — a² + b².z · z̄ = (a+bi)(a−bi) = a² − (bi)² = a² − b²i² = a² + b². This is always a non-negative real number.
What is i^100? — 1.100 = 4·25, so i^100 = (i^4)^25 = 1^25 = 1. Any power of i that is divisible by 4 equals 1.
What is the sum of a complex number and its conjugate? — 2a (real number).(a+bi) + (a−bi) = 2a, which is always a real number. The imaginary parts cancel.
What is the difference of a complex number and its conjugate (z − z̄)? — 2bi.(a+bi) − (a−bi) = 2bi, which is purely imaginary. The real parts cancel.
Given z₁ = 2 + 3i and z₂ = 1 − i, compute z₁z₂. — 5 + i.(2+3i)(1−i) = 2−2i+3i−3i² = 2+i−3(−1) = 2+i+3 = 5+i.
What is the sum of the roots of x² + 6x + 25 = 0? — −6.By Vieta's formulas, sum of roots = −b/a = −6/1 = −6.
What is the product of the roots of x² + 6x + 25 = 0? — 25.By Vieta's formulas, product of roots = c/a = 25/1 = 25.
Find the solutions to x² + 4x + 8 = 0. — x = −2 ± 2i.Discriminant = 16 − 32 = −16. x = (−4 ± √(−16))/2 = (−4 ± 4i)/2 = −2 ± 2i.
Solve 3x² + 12 = 0. — x = ±2i.3x² = −12 → x² = −4 → x = ±2i.
Which statement is always true about the complex roots of a quadratic with real coefficients? — They are conjugates of each other.Complex roots of polynomials with real coefficients always come in conjugate pairs. This follows from the fact that if z is a root, so is z̄.
A quadratic equation has discriminant −100. What are its roots if a = 1, b = 4? — x = −2 ± 5i.x = (−4 ± √(−100))/2 = (−4 ± 10i)/2 = −2 ± 5i.
The product of complex conjugate roots a ± bi is always: — a² + b².(a+bi)(a−bi) = a² − (bi)² = a² − b²(−1) = a² + b². The product of conjugates is always a² + b², a non-negative real number.
What does the Fundamental Theorem of Algebra guarantee for a degree-4 polynomial? — Exactly 4 roots in ℂ.The Fundamental Theorem of Algebra states every degree-n polynomial has exactly n roots in ℂ (counting multiplicity). For degree 4, there are exactly 4 complex roots.
What is the argument of the complex number 1 + i (in radians)? — π/4.arg(1 + i) = arctan(1/1) = arctan(1) = π/4 (since both parts are equal and positive, it's in Q1 at 45°).
By what angle does multiplying a complex number by i rotate it in the complex plane? — 90° counterclockwise.Multiplying by i rotates a complex number 90° counterclockwise (since i = e^(iπ/2), which represents a 90° rotation).
What is the polar form of a complex number with modulus r and argument θ? — r(cosθ + i sinθ).The polar form is z = r(cosθ + i sinθ), where r = |z| is the modulus and θ = arg(z) is the argument.
What is the modulus of the complex number 8 + 15i? — 17.|8 + 15i| = √(64 + 225) = √289 = 17. Note: √289 = 17 is also correct (same answer as choice 1 and 2).
Adding two complex numbers in the complex plane follows the same rule as: — Vector addition (parallelogram rule).Adding complex numbers in the plane follows the parallelogram rule for vector addition, since complex numbers can be treated as 2D vectors.
What is the modulus of the product z₁z₂ if |z₁| = 3 and |z₂| = 4? — 12.|z₁z₂| = |z₁| · |z₂| = 3 · 4 = 12. The modulus of a product equals the product of the moduli.
What is the argument of a negative real number, e.g., −5? — π.Negative real numbers lie on the negative real axis, making an angle of π (180°) with the positive real axis.
What is |z̄| if |z| = 7? — 7.The modulus of the conjugate equals the modulus of the original: |z̄| = |a−bi| = √(a²+b²) = |z|. So |z̄| = 7.
What is the argument of z = −1 + 0i? — π.The complex number −1 lies on the negative real axis. Its argument is π (180°), the standard angle for negative real numbers.
If |z| = 0, what can we conclude about z? — z = 0.|z| = √(a² + b²) = 0 implies a² + b² = 0, which means a = 0 and b = 0. Therefore z = 0.