Complex numbers (Algebra II) · lesson

The imaginary unit i

💡 i = √(−1), so the square root of a negative number is a real square root times i.

No real number squared gives a negative. So mathematicians made a new number, i, with i² = −1.

√(−36) = √36 × √(−1) = 6i. A complex number has a real part and an imaginary part, like −2 + 9i.

🧠 Let's work it out

Write −2 + √(−81) as a complex number.

√(−81) = √81 × i = 9i.
The number is −2 + 9i.
The real part is −2 and the imaginary part is 9.

What do you think comes next?

imaginary unit: i, the number whose square is −1
complex number: a number a + bi with a real part a and an imaginary part b

🔗 Khan Academy: Complex numbers · Khan Academy (free) ↗

Adding and subtracting

💡 Combine real parts with real parts and imaginary parts with imaginary parts.

Treat i like a variable. (6 − 2i) + (−1 + 5i) = 5 + 3i.

When you subtract, the minus sign changes both parts of the second number.

🧠 Let's work it out

Find (7 + i) − (3 − 4i).

Change both signs in the second number: 7 + i − 3 + 4i.
Real parts: 7 − 3 = 4.
Imaginary parts: i + 4i = 5i.
So the answer is 4 + 5i.

What do you think comes next?

real part: the number without i, like 4 in 4 + 5i
imaginary part: the number in front of i, like 5 in 4 + 5i

Multiplying complex numbers

💡 Multiply like binomials, then replace every i² with −1.

Use FOIL or distribute, just as with (x + 3)(x − 2). Every product of two i terms makes an i².

Swap i² for −1 and combine. A cube like (a + bi)³ is easiest one step at a time: square it, then multiply once more.

🧠 Let's work it out

Multiply (3 + i)(2 − 4i).

FOIL: 6 − 12i + 2i − 4i².
Combine the i terms: 6 − 10i − 4i².
Replace i² with −1: 6 − 10i + 4.
Combine the real parts: 10 − 10i.

What do you think comes next?

FOIL: first, outer, inner, last: the four products when multiplying two binomials

Powers of i repeat

💡 Powers of i cycle every 4: i, −1, −i, 1, so only the remainder after dividing by 4 matters.

i¹ = i, i² = −1, i³ = −i, and i⁴ = 1. Then the pattern starts over.

To find a big power, divide the exponent by 4. The remainder tells you where you are in the cycle. A remainder of 0 means the answer is 1.

🧠 Let's work it out

Find i¹⁷.

17 = 4 × 4 + 1, so the remainder is 1.
So i¹⁷ = i¹ = i.
Write it as a + bi: i = 0 + 1i.

What do you think comes next?

exponent: the small raised number that says how many times to multiply

Conjugates and division

💡 A number times its conjugate is real, so multiply top and bottom by the bottom's conjugate to divide.

The conjugate of a + bi is a − bi. Their product is a² + b², a real number. The i terms cancel.

To divide, multiply the top and bottom by the conjugate of the bottom. The bottom becomes real. Then divide each part by it.

🧠 Let's work it out

Divide (4 + 2i) by (1 − i).

The conjugate of 1 − i is 1 + i.
Multiply the top by it: (4 + 2i)(1 + i) = 4 + 4i + 2i + 2i² = 2 + 6i.
Multiply the bottom by it: (1 − i)(1 + i) = 1 + 1 = 2.
(2 + 6i) ÷ 2 = 1 + 3i.

What do you think comes next?

conjugate: the same complex number with the sign of the imaginary part flipped

Quadratics with complex solutions

💡 When the discriminant is negative, the quadratic formula gives two complex solutions a ± bi.

The quadratic formula is x = (−b ± √(b² − 4ac)) ÷ 2a. The part under the root, b² − 4ac, is the discriminant.

If the discriminant is negative, its square root has an i. The two answers are conjugates: one with + bi and one with − bi.

🧠 Let's work it out

Solve x² − 6x + 13 = 0.

Discriminant = 36 − 52 = −16.
Its square root: √(−16) = 4i.
Use the formula: x = (6 ± 4i) ÷ 2 = 3 ± 2i.
The solution with a positive imaginary part is 3 + 2i.

What do you think comes next?

discriminant: b² − 4ac, the part under the square root in the quadratic formula
quadratic formula: x = (−b ± √(b² − 4ac)) ÷ 2a