Grade: Grade 11 Subject: Mathematics Unit: Algebra II Completion SAT: AdvancedMath ACT: Math

Complex Numbers

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What Are Complex Numbers?

Complex numbers extend our number system beyond real numbers. They were invented to solve equations like x² + 1 = 0, which has no solution in real numbers because no real number squared gives a negative result.

Definition: The Imaginary Unit

The imaginary unit, denoted i, is defined as:

i = √(-1), which means i² = -1

Definition: Complex Number

A complex number is a number of the form:

z = a + bi

where a is the real part and b is the imaginary part, and both a and b are real numbers.

The Complex Number System

  • Real numbers are complex numbers where b = 0 (e.g., 5 = 5 + 0i)
  • Pure imaginary numbers are complex numbers where a = 0 (e.g., 3i = 0 + 3i)
  • Complex numbers have both nonzero real and imaginary parts (e.g., 2 + 3i)

Operations with Complex Numbers

Addition and Subtraction

Add or subtract the real parts and imaginary parts separately:

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Multiplication

Use the distributive property (FOIL) and remember that i² = -1:

(a + bi)(c + di) = ac + adi + bci + bdi²

= ac + adi + bci + bd(-1)

= (ac - bd) + (ad + bc)i

Powers of i

The powers of i follow a cycle of 4:

  • i¹ = i
  • i² = -1
  • i³ = i² · i = -1 · i = -i
  • i⁴ = i² · i² = (-1)(-1) = 1
  • i⁵ = i⁴ · i = 1 · i = i (cycle repeats)

Tip: To find i^n, divide n by 4 and use the remainder to determine the result.

Complex Conjugates

Definition: Complex Conjugate

The complex conjugate of z = a + bi is:

z̄ = a - bi

The conjugate has the same real part but the opposite sign on the imaginary part.

Key Property: When you multiply a complex number by its conjugate, you always get a real number:

(a + bi)(a - bi) = a² - (bi)² = a² - b²i² = a² - b²(-1) = a² + b²

Division

To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator:

(a + bi)/(c + di) = [(a + bi)(c - di)]/[(c + di)(c - di)] = [(a + bi)(c - di)]/(c² + d²)

Examples

Example 1: Adding Complex Numbers

Problem: Simplify (3 + 4i) + (2 - 7i)

Solution:

Step 1: Group real parts and imaginary parts

= (3 + 2) + (4i + (-7i))

Step 2: Simplify each group

= 5 + (-3i)

= 5 - 3i

Example 2: Multiplying Complex Numbers

Problem: Simplify (2 + 3i)(4 - 5i)

Solution:

Step 1: Use FOIL (First, Outer, Inner, Last)

= (2)(4) + (2)(-5i) + (3i)(4) + (3i)(-5i)

= 8 - 10i + 12i - 15i²

Step 2: Replace i² with -1

= 8 - 10i + 12i - 15(-1)

= 8 - 10i + 12i + 15

Step 3: Combine like terms

= (8 + 15) + (-10i + 12i)

= 23 + 2i

Example 3: Powers of i

Problem: Simplify i⁴⁷

Solution:

Step 1: Divide the exponent by 4

47 ÷ 4 = 11 remainder 3

Step 2: The remainder tells us which power in the cycle

i⁴⁷ = i³ = -i

Example 4: Division of Complex Numbers

Problem: Simplify (3 + 2i)/(1 - 4i)

Solution:

Step 1: Multiply numerator and denominator by the conjugate of the denominator (1 + 4i)

= [(3 + 2i)(1 + 4i)]/[(1 - 4i)(1 + 4i)]

Step 2: Expand the numerator using FOIL

= (3 + 12i + 2i + 8i²)/(1 - 16i²)

= (3 + 14i + 8(-1))/(1 - 16(-1))

= (3 + 14i - 8)/(1 + 16)

= (-5 + 14i)/17

Step 3: Write in standard form

= -5/17 + (14/17)i

Example 5: Product of Conjugates

Problem: Find the product of 3 - 5i and its conjugate.

Solution:

Step 1: Identify the conjugate of 3 - 5i

Conjugate = 3 + 5i

Step 2: Multiply using the formula (a + bi)(a - bi) = a² + b²

(3 - 5i)(3 + 5i) = 3² + 5² = 9 + 25 = 34

Practice

Solve these problems. Answers are provided below for self-checking.

1. Simplify: (5 + 2i) + (-3 + 6i)

2. Simplify: (7 - 4i) - (2 - 9i)

3. Simplify: (1 + i)(1 - i)

4. Simplify: (2 - 3i)²

5. Simplify: i²⁵

6. Simplify: i¹⁰⁰

7. Find the conjugate of -4 + 7i and verify by finding their product.

8. Simplify: (4 + i)/(2 - i)

9. Simplify: 6i/(3 + i)

10. Solve for x and y if (2x + 3yi) + (4 - 2i) = 10 + 7i

Click to reveal answers
  1. 2 + 8i
  2. 5 + 5i
  3. 2 (a real number!)
  4. -5 - 12i
  5. i (since 25 ÷ 4 = 6 R 1)
  6. 1 (since 100 ÷ 4 = 25 R 0, so i¹⁰⁰ = i⁴ = 1)
  7. Conjugate: -4 - 7i; Product: (-4)² + 7² = 16 + 49 = 65
  8. (7 + 6i)/5 or 7/5 + (6/5)i
  9. (3 + 9i)/5 or 3/5 + (9/5)i
  10. x = 3, y = 3

Check Your Understanding

1. Why do we need complex numbers? What types of equations can we solve with complex numbers that we couldn't solve with just real numbers?

Show answer

Complex numbers allow us to solve polynomial equations that have no real solutions. For example, x² + 1 = 0 has solutions x = i and x = -i. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n roots (counting multiplicity) in the complex number system.

2. Explain why i² = -1 but √(-1) · √(-1) ≠ √((-1)(-1)) = √1 = 1.

Show answer

The rule √a · √b = √(ab) only works when both a and b are non-negative. When dealing with negative numbers under square roots, we must first express them using i and then simplify. So √(-1) · √(-1) = i · i = i² = -1, not 1.

3. What is special about multiplying a complex number by its conjugate?

Show answer

When you multiply a complex number by its conjugate, the result is always a non-negative real number: (a + bi)(a - bi) = a² + b². This property is essential for dividing complex numbers, as it allows us to create a real number in the denominator.

4. How can you quickly determine i^n for any positive integer n?

Show answer

Divide n by 4 and look at the remainder: if remainder is 0, i^n = 1; if remainder is 1, i^n = i; if remainder is 2, i^n = -1; if remainder is 3, i^n = -i. This works because the powers of i cycle with period 4.

🚀 Next Steps

  • Review any concepts that felt challenging
  • Move on to the next lesson when ready
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