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Learn · Algebra 2

Complex Numbers: When the Quadratic Formula Hits a Negative Discriminant

The imaginary unit i = √(−1) extends the number line into a plane. Master arithmetic with complex numbers, the conjugate trick for division, and you will handle every "no real solution" quadratic on the CBE.

8 minTEKS 4FAlgebra 2

When real numbers run out, math invents new ones

The equation x² = −1 has no real solution — there's no real number that squared gives a negative. So mathematicians invented one: i, defined so that i² = −1. Once you accept i, every quadratic has exactly two roots, and the entire algebraic universe stays consistent.

Big idea

Complex numbers have the form a + bi: a real part and an imaginary part. Treat i like a variable, with the rule i² = −1. Everything else follows.

The defining equation
What is i², where i is the imaginary unit?
Complex plane: z = 3 + 4i has magnitude |z| = √(3² + 4²) = 5 (3-4-5 triangle!)ReIm1231i2i3i4iz = 3 + 4iRe(z) = 3Im(z) = 4|z| = 5|z| = √(Re² + Im²) = √(9 + 16) = √25 = 5The magnitude is the distance from origin to the point (Pythagorean theorem)

Powers of i — the cycle of four

i1 = ii2 = −1i3 = i2 · i = −ii4 = (i2)2 = 1 (then it repeats)Divide the exponent by 4 and look at the remainder: 0 → 1, 1 → i, 2 → −1, 3 → −i.

Arithmetic with complex numbers

Add / subtract
Combine real with real, imaginary with imaginary. (3 + 4i) + (1 − 2i) = 4 + 2i.
Multiply
Distribute (FOIL), then replace i² with −1. (1 + i)(2 + 3i) = 2 + 3i + 2i + 3i² = 2 + 5i − 3 = −1 + 5i.
Conjugate
Flip the sign of the imaginary part: conjugate of (a + bi) is (a − bi). Used to simplify division.
Powers of i cycle every 4: {i, −1, −i, 1, i, ...}i¹= ii²= −1i³= −ii⁴= 1cycles back to i¹

Division: multiply by the conjugate

To simplify a fraction with i in the denominator, multiply top and bottom by the conjugate of the denominator. The result is a real denominator.

(1 + i) / (1 − i)= (1 + i)(1 + i) / [(1 − i)(1 + i)] (multiply by conjugate)= (1 + 2i + i²) / (1 − i²)= (1 + 2i − 1) / (1 + 1) = 2i / 2= i

When the discriminant is negative

From the previous lesson: discriminant < 0 means no real roots. With complex numbers, those roots exist — they're just imaginary.

x² + 4x + 5 = 0 (discriminant = 16 − 20 = −4)x = (−4 ± √−4) / 2√−4 = √4 · √−1 = 2ix = (−4 ± 2i) / 2 = −2 ± iComplex roots always come in conjugate pairs (a + bi and a − bi) for polynomials with real coefficients.
Conjugate z̄ = a − bi: reflection over real axis; z·z̄ = a² + b² (real)z = 3 + 2iz̄ = 3 − 2iz · z̄ = (3+2i)(3−2i) = 9 + 4 = 13 (real!)

3-second recap

  • i = √(−1), so i² = −1. The defining equation.
  • Add/subtract: combine matching parts. Multiply: FOIL, then replace i².
  • Divide: multiply by the conjugate of the denominator.
  • Negative discriminant → complex roots in conjugate pairs.

Check yourself

Quick check #1
Simplify: (3 + 2i) + (4 − 5i).
Quick check #2
What is i² ?