Every operation on complex numbers is ordinary algebra with exactly one extra fact available: \(i^{2}=-1\). This lesson works through addition, subtraction, multiplication, and division on complex numbers of the form \(a+bi\), treating the real part and the imaginary part the way Adding Polynomials and Subtracting Polynomials treat like terms — because that’s really what they are.
Adding and Subtracting Complex Numbers
Combine the real parts together and the imaginary parts together, exactly as like terms combine in a polynomial — a real number and an imaginary number are never like terms with each other.
$$ (3+5i) + (2-4i) = (3+2) + (5-4)i = 5 + i $$
$$ (7-2i) - (4+6i) = (7-4) + (-2-6)i = 3 - 8i $$
Subtraction distributes the negative sign across both parts of the second complex number, exactly the same way it distributes across every term of a polynomial.
Multiplying Complex Numbers
A complex number \(a+bi\) is a binomial with \(i\) standing in for a variable, so multiplying two of them uses ordinary FOIL — with one extra simplification step at the end, since \(i^{2}=-1\).
$$ (3+2i)(1+4i) $$
$$ = 3(1) + 3(4i) + 2i(1) + 2i(4i) = 3 + 12i + 2i + 8i^{2} $$
$$ = 3 + 14i + 8(-1) = 3 + 14i - 8 = -5 + 14i $$
The \(i^{2}\) term is the one genuinely new thing that happens compared to multiplying two ordinary binomials — every \(i^{2}\) that appears gets replaced with \(-1\) before the real and imaginary parts are combined.
Powers of i
The powers of \(i\) repeat in a cycle of four:
| Power | Value |
|---|---|
| \(i^{1}\) | \(i\) |
| \(i^{2}\) | \(-1\) |
| \(i^{3}\) | \(-i\) |
| \(i^{4}\) | \(1\) |
$$ i^{5} = i^{4}\cdot i = 1 \cdot i = i \qquad \text{(the cycle simply restarts)} $$
For any power, divide the exponent by \(4\) and match the remainder to the table: a remainder of \(1\) behaves like \(i^{1}\), a remainder of \(2\) like \(i^{2}\), and so on, with a remainder of \(0\) behaving like \(i^{4}=1\).
Dividing Complex Numbers
Since \(i\) can’t be left in a denominator, dividing by a complex number multiplies the numerator and denominator by the denominator’s conjugate — the same complex number with the sign of its imaginary part flipped.
$$ \frac{1}{3+2i} \cdot \frac{3-2i}{3-2i} = \frac{3-2i}{9-4i^{2}} = \frac{3-2i}{9-4(-1)} = \frac{3-2i}{13} $$
The denominator \((3+2i)(3-2i)\) always simplifies to \(a^{2}+b^{2}\) — a real number — because it’s a difference of squares where the “squared” term \(i^{2}\) becomes \(-1\) and flips the subtraction into addition.
Worked Example A: Adding and Subtracting
Simplify \((6-3i) + (2+7i) - (4-i)\).
Step 1 — combine the first two complex numbers:
$$ (6+2) + (-3+7)i = 8 + 4i $$
Step 2 — subtract the third, distributing the negative sign to both of its parts:
$$ (8-4) + (4-(-1))i = 4 + 5i $$
Answer: \(4+5i\)
Worked Example B: Multiplying Two Complex Numbers
Multiply \((4-i)(2+3i)\).
Step 1 — apply FOIL:
$$ 4(2) + 4(3i) + (-i)(2) + (-i)(3i) = 8 + 12i - 2i - 3i^{2} $$
Step 2 — replace \(i^{2}\) with \(-1\):
$$ 8 + 10i - 3(-1) = 8 + 10i + 3 $$
Step 3 — combine the real parts:
$$ 11 + 10i $$
Answer: \(11+10i\)
Worked Example C: A Higher Power of i
Simplify \(i^{23}\).
Step 1 — divide the exponent by \(4\) and find the remainder:
$$ 23 = 4(5) + 3 $$
Step 2 — match the remainder \(3\) to the power-of-\(i\) table:
$$ i^{23} = i^{3} = -i $$
Answer: \(-i\)
Worked Example D: Dividing Complex Numbers
Divide \(\dfrac{5+i}{2-3i}\).
Step 1 — multiply the numerator and denominator by the conjugate of the denominator:
$$ \frac{5+i}{2-3i} \cdot \frac{2+3i}{2+3i} $$
Step 2 — expand the denominator using the difference of squares pattern:
$$ (2-3i)(2+3i) = 4 - 9i^{2} = 4 - 9(-1) = 13 $$
Step 3 — expand the numerator with FOIL:
$$ (5+i)(2+3i) = 10 + 15i + 2i + 3i^{2} = 10 + 17i + 3(-1) = 7 + 17i $$
Answer: \(\dfrac{7+17i}{13} = \dfrac{7}{13} + \dfrac{17}{13}i\)
Common Mistakes to Avoid
- Combining real and imaginary parts as if they were like terms with each other. \(3+5i\) cannot be simplified to \(8i\) or to \(8\) — the two parts stay separate unless another real or imaginary term is being added to the matching part.
- Forgetting to replace \(i^{2}\) with \(-1\) after multiplying. Leaving a term like \(8i^{2}\) unsimplified is not a finished answer; it must become \(-8\).
- Losing a sign when \(i^{2}\) turns a subtraction into an addition. \(9-4i^{2}\) becomes \(9-4(-1)=9+4=13\), not \(9-4=5\) — the double negative is easy to drop by accident.
- Multiplying only the denominator by the conjugate, not the numerator. Both parts of the fraction need the same multiplier, since it has to equal \(1\) to avoid changing the fraction’s value.
- Writing the conjugate with the wrong sign. The conjugate of \(2-3i\) is \(2+3i\) — only the imaginary part’s sign flips, never the real part’s.
- Miscounting the cycle when evaluating a power of \(i\). The cycle length is \(4\), not \(2\) or \(3\); dividing the exponent by \(4\) and using the remainder is the reliable way to avoid a miscount.
Where This Shows Up Later
- The quadratic formula. A negative discriminant produces two complex solutions, and reporting them in \(a+bi\) form uses exactly this lesson’s simplification rules.
- Rationalizing Denominators. The conjugate trick used here to clear \(i\) from a denominator is the identical technique used there to clear a radical.
- Polar and trigonometric form of complex numbers. Multiplying and dividing complex numbers in polar form later gives a much faster alternative to FOIL, once angles and magnitudes are introduced.
- Electrical engineering and signal processing. Complex number arithmetic, exactly as covered here, is the standard tool for combining alternating-current quantities in those fields.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Simplify \((4+3i) + (1-6i)\).
Show answer
$$ (4+1) + (3-6)i = 5 - 3i $$
Answer: \(5-3i\)
Problem 2. Simplify \((8-2i) - (3-5i)\).
Show answer
$$ (8-3) + (-2-(-5))i = 5 + 3i $$
Answer: \(5+3i\)
Problem 3. Multiply \((2+i)(3-i)\).
Show answer
$$ 2(3) + 2(-i) + i(3) + i(-i) = 6 - 2i + 3i - i^{2} = 6 + i - (-1) = 7 + i $$
Answer: \(7+i\)
Problem 4. Simplify \(i^{10}\).
Show answer
$$ 10 = 4(2)+2 \;\Longrightarrow\; i^{10} = i^{2} = -1 $$
Answer: \(-1\)
Problem 5. Multiply \((5-2i)(5+2i)\).
Show answer
Difference of squares pattern:
$$ 25 - 4i^{2} = 25 - 4(-1) = 29 $$
Answer: \(29\) (a real number, since these two complex numbers are conjugates of each other)
Problem 6. Divide \(\dfrac{1}{4+i}\).
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Step 1 — multiply by the conjugate:
$$ \frac{1}{4+i} \cdot \frac{4-i}{4-i} $$
Step 2 — expand the denominator:
$$ 16 - i^{2} = 16-(-1) = 17 $$
Answer: \(\dfrac{4-i}{17} = \dfrac{4}{17} - \dfrac{1}{17}i\)
Problem 7. Simplify \((3+2i) + (-1+2i) - (2-4i)\).
Show answer
Step 1 — combine the first two:
$$ (3-1) + (2+2)i = 2 + 4i $$
Step 2 — subtract the third:
$$ (2-2) + (4-(-4))i = 0 + 8i = 8i $$
Answer: \(8i\)
Problem 8. Multiply \((1-3i)^{2}\).
Show answer
Step 1 — apply the perfect-square pattern (or FOIL directly):
$$ 1^{2} - 2(1)(3i) + (3i)^{2} = 1 - 6i + 9i^{2} $$
Step 2 — replace \(i^{2}\) with \(-1\):
$$ 1 - 6i + 9(-1) = 1 - 6i - 9 = -8 - 6i $$
Answer: \(-8-6i\)
Problem 9. Divide \(\dfrac{3-4i}{1+2i}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{3-4i}{1+2i} \cdot \frac{1-2i}{1-2i} $$
Step 2 — expand the denominator:
$$ 1 - 4i^{2} = 1-4(-1) = 5 $$
Step 3 — expand the numerator:
$$ (3-4i)(1-2i) = 3 - 6i - 4i + 8i^{2} = 3 - 10i + 8(-1) = -5 - 10i $$
Answer: \(\dfrac{-5-10i}{5} = -1-2i\)
Problem 10. Simplify \(i^{47}\).
Show answer
$$ 47 = 4(11) + 3 \;\Longrightarrow\; i^{47} = i^{3} = -i $$
Answer: \(-i\)
Quick Reference
| Operation | Move |
|---|---|
| Add or subtract | Combine real parts together and imaginary parts together, separately |
| Multiply | FOIL, then replace every \(i^{2}\) with \(-1\) and combine like parts |
| Divide | Multiply top and bottom by the denominator’s conjugate |
| Conjugate of \(a+bi\) | \(a-bi\) |
| Complex number times its conjugate | Always \(a^{2}+b^{2}\), a real number |
| Any power of \(i\) | Divide the exponent by \(4\); match the remainder to \(i^{1}, i^{2}, i^{3}, i^{4}\) |
Every technique here reuses Adding Polynomials, Multiplying Polynomials, and the conjugate trick from Rationalizing Denominators — complex numbers just add the single fact \(i^{2}=-1\) on top. Revisit Complex Numbers for where \(i\) comes from and how it extends the real number system. Build speed with the Complex Number Arithmetic Generator, or browse the rest of the Algebra lessons as new ones publish.