Operations with Complex Numbers

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Operations with complex numbers work like operations with binomials. To add or subtract, combine the real parts and combine the imaginary parts. To multiply, use the distributive property or FOIL, then replace i^2 with −1. Always write the answer in standard form, a + bi.

Before this lesson: You should know Complex Numbers, Combining Like Terms, Multiplying Binomials and Special Products of Binomials.

Treat i Like a Variable, Then Simplify

Think of i as a variable, like x. Then 3 + 5i works like 3 + 5x. You combine like terms and multiply binomials the same way.

There is one extra step. Whenever i^2 appears, replace it with −1. That turns the i^2 term into a real number.

Rules for Adding, Subtracting and Multiplying

Adding and subtracting

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

Adding complex numbers looks like adding arrows in the complex plane. Place the second arrow at the tip of the first. The sum is where you end up.

Complex plane with three arrows from the origin. A blue arrow goes to 2 + 3i and a green arrow goes to 4 − i. A dashed green copy of the 4 − i arrow starts at the tip of 2 + 3i and ends at 6 + 2i. An orange arrow from the origin to 6 + 2i shows the sum: (2 + 3i) + (4 − i) = 6 + 2i.
Adding 4 − i to 2 + 3i moves 4 right and 1 down, landing on 6 + 2i.

Multiplying

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

You don’t need to memorize this formula. Just use FOIL and replace i^2 with −1.

Complex conjugates

The complex conjugate of a + bi is a - bi. Only the sign of the imaginary part changes. A number times its conjugate is always a real number.

(a + bi)(a - bi) = a^2 + b^2

Worked Examples

Example 1: Add and subtract

Simplify (a) (3 + 5i) + (2 - 8i) and (b) (7 - 2i) - (4 - 6i).

  1. (a) Group the real parts and the imaginary parts (see combining like terms): (3 + 2) + (5 - 8)i
  2. Add: 5 - 3i
  3. (b) Distribute the minus sign to both parts: 7 - 2i - 4 + 6i
  4. Group the real parts and the imaginary parts: (7 - 4) + (-2 + 6)i
  5. Simplify: 3 + 4i

Answer: (a) 5 - 3i (b) 3 + 4i

Example 2: Multiply by an imaginary number

Simplify 3i(4 - 2i).

  1. Distribute 3i (see distributive property): 3i \cdot 4 - 3i \cdot 2i
  2. Multiply: 12i - 6i^2
  3. Replace i^2 with −1: 12i - 6(-1)
  4. Simplify: 12i + 6
  5. Write in standard form: 6 + 12i

Answer: 6 + 12i

Example 3: Multiply two complex numbers

Simplify (2 + 3i)(4 - i).

  1. Use FOIL (see multiplying binomials): 2 \cdot 4 + 2(-i) + 3i \cdot 4 + 3i(-i)
  2. Multiply each pair: 8 - 2i + 12i - 3i^2
  3. Combine the i terms: 8 + 10i - 3i^2
  4. Replace i^2 with −1: 8 + 10i - 3(-1)
  5. Simplify: 8 + 10i + 3
  6. Combine the real parts: 11 + 10i

Answer: 11 + 10i

Example 4: Conjugates and squares

Simplify (a) (5 + 2i)(5 - 2i) and (b) (3 - 4i)^2.

  1. (a) The two numbers are conjugates. FOIL: 25 - 10i + 10i - 4i^2
  2. The i terms cancel: 25 - 4i^2
  3. Replace i^2 with −1: 25 - 4(-1)
  4. Simplify: 25 + 4 = 29
  5. (b) Write the square as a product: (3 - 4i)(3 - 4i)
  6. FOIL: 9 - 12i - 12i + 16i^2
  7. Combine the i terms: 9 - 24i + 16i^2
  8. Replace i^2 with −1: 9 - 24i - 16
  9. Combine the real parts: -7 - 24i

Answer: (a) 29 (b) -7 - 24i

Part (a) checks the conjugate rule: 5^2 + 2^2 = 25 + 4 = 29.

Example 5: Word problem with an electric circuit

In an AC circuit, two parts in a row (in series) have impedances Z_1 = 4 + 3i ohms and Z_2 = 6 - 8i ohms. The total impedance is their sum. The voltage is V = IZ, where I is the current. If the current is I = 2 - i amps, find the total impedance and the voltage.

  1. Add the impedances: Z = (4 + 3i) + (6 - 8i)
  2. Group the parts: (4 + 6) + (3 - 8)i
  3. Simplify: Z = 10 - 5i ohms
  4. Write the voltage formula: V = (2 - i)(10 - 5i)
  5. FOIL: 20 - 10i - 10i + 5i^2
  6. Combine the i terms: 20 - 20i + 5i^2
  7. Replace i^2 with −1: 20 - 20i - 5
  8. Combine the real parts: 15 - 20i

Answer: The total impedance is 10 - 5i ohms, and the voltage is 15 - 20i volts.

Common Mistakes

MistakeExample of the errorFix
Not distributing the minus sign(7 - 2i) - (4 - 6i) = 3 - 8iSubtract both parts: -(4 - 6i) = -4 + 6i, so the answer is 3 + 4i.
Leaving i^2 in the answer8 + 10i - 3i^2Replace i^2 with −1 and combine: 11 + 10i.
Replacing i^2 with 1-3i^2 = -3-3i^2 = -3(-1) = 3
Squaring each part separately(3 + 4i)^2 = 9 + 16i^2 = -7Use FOIL. The middle term 24i is missing: (3 + 4i)^2 = -7 + 24i.
Combining real and imaginary parts5 - 3i = 2iReal and imaginary parts are unlike terms. 5 - 3i is already simplified.

Practice Problems

  1. Simplify (6 + 2i) + (-1 + 7i).
  2. Simplify (8 - 3i) - (2 + 5i).
  3. Simplify (-4 + i) - (-9 - 6i).
  4. Simplify 2i(5 + 3i).
  5. Simplify (1 + 4i)(3 - 2i).
  6. Simplify (6 - i)(2 + 5i).
  7. Simplify (4 + 7i)(4 - 7i).
  8. Simplify (2 + 5i)^2.
  9. Simplify \sqrt{-9} \cdot \sqrt{-16}.
  10. Two circuit parts in series have impedances 3 + 2i and 5 - 7i ohms. The current is 1 + 2i amps. Find the total impedance and the voltage V = IZ.
Show answers
  1. Group: (6 - 1) + (2 + 7)i
    Answer: 5 + 9i
  2. Distribute the minus sign: 8 - 3i - 2 - 5i
    Group: (8 - 2) + (-3 - 5)i
    Answer: 6 - 8i
  3. Distribute the minus sign: -4 + i + 9 + 6i
    Group: (-4 + 9) + (1 + 6)i
    Answer: 5 + 7i
  4. Distribute: 10i + 6i^2
    Replace i^2: 10i - 6
    Answer: -6 + 10i
  5. FOIL: 3 - 2i + 12i - 8i^2
    Combine i terms: 3 + 10i - 8i^2
    Replace i^2: 3 + 10i + 8
    Answer: 11 + 10i
  6. FOIL: 12 + 30i - 2i - 5i^2
    Combine i terms: 12 + 28i - 5i^2
    Replace i^2: 12 + 28i + 5
    Answer: 17 + 28i
  7. FOIL: 16 - 28i + 28i - 49i^2
    The i terms cancel: 16 - 49i^2
    Replace i^2: 16 + 49
    Answer: 65
  8. Write as a product: (2 + 5i)(2 + 5i)
    FOIL: 4 + 10i + 10i + 25i^2
    Combine i terms: 4 + 20i + 25i^2
    Replace i^2: 4 + 20i - 25
    Answer: -21 + 20i
  9. Write each root with i first: \sqrt{-9} = 3i and \sqrt{-16} = 4i
    Multiply: 3i \cdot 4i = 12i^2
    Replace i^2: 12(-1) = -12
  10. Add: (3 + 5) + (2 - 7)i = 8 - 5i ohms
    Voltage: V = (1 + 2i)(8 - 5i)
    FOIL: 8 - 5i + 16i - 10i^2
    Combine i terms: 8 + 11i - 10i^2
    Replace i^2: 8 + 11i + 10
    Answer: 18 + 11i volts

FAQ

How do you add complex numbers?

Add the real parts together and add the imaginary parts together. For example, (2 + 3i) + (4 - i) = 6 + 2i.

How do you multiply complex numbers?

Multiply them like binomials, using FOIL. Then replace i^2 with −1 and combine like terms.

What is the conjugate of a complex number?

The conjugate of a + bi is a - bi. Their product, a^2 + b^2, is always a real number. The next lesson, Dividing Complex Numbers, uses this to divide complex numbers.

Why does i squared equal −1?

That is how i is defined. It is the number whose square is −1, so i^2 = -1 by definition.

Summary

  • To add or subtract complex numbers, combine the real parts and combine the imaginary parts.
  • Distribute the minus sign to both parts when you subtract.
  • To multiply, use the distributive property or FOIL, then replace i^2 with −1.
  • The conjugate of a + bi is a - bi, and (a + bi)(a - bi) = a^2 + b^2.
  • Write every answer in standard form, a + bi.

Previous lesson: Complex Numbers

Next lesson: Dividing Complex Numbers

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