Radicals stops just short of one question: what is \(\sqrt{-1}\)? No real number answers it, since no real number squared gives a negative result. Complex numbers exist to answer that question directly, and the conjugate trick that rationalizes a radical denominator turns out to be exactly the tool that divides complex numbers too.
What Is a Complex Number?
Define \(i\), the imaginary unit, as the number whose square is \(-1\):
$$ i^{2} = -1, \qquad \text{equivalently} \qquad i = \sqrt{-1} $$
A complex number has the standard form \(a+bi\), where \(a\) is the real part and \(b\) is the imaginary part (both ordinary real numbers). \(3+2i\), \(-1+i\), and \(5\) (where \(b=0\)) are all complex numbers — every real number is a complex number with imaginary part zero.
Powers of i: A Repeating Cycle
Powers of \(i\) repeat in a cycle of four, which follows directly from \(i^{2}=-1\):
$$ i^{1}=i, \qquad i^{2}=-1, \qquad i^{3}=i^{2} \cdot i=-i, \qquad i^{4}=i^{2} \cdot i^{2}=(-1)(-1)=1 $$
Since \(i^{4}=1\), multiplying by \(i^{4}\) again just returns to \(i^{5}=i\), and the pattern repeats forever. To simplify any power of \(i\), divide the exponent by \(4\) and use the remainder: \(0 \to 1\), \(1 \to i\), \(2 \to -1\), \(3 \to -i\).
Properties of Complex Numbers
| Rule | Statement |
|---|---|
| Definition | \(i^{2} = -1\) |
| Addition | \((a+bi)+(c+di) = (a+c)+(b+d)i\) |
| Subtraction | \((a+bi)-(c+di) = (a-c)+(b-d)i\) |
| Multiplication | \((a+bi)(c+di) = (ac-bd)+(ad+bc)i\) |
| Complex Conjugate | The conjugate of \(a+bi\) is \(a-bi\) |
| Conjugate Product | \((a+bi)(a-bi) = a^{2}+b^{2}\) |
Addition and subtraction combine real parts with real parts and imaginary parts with imaginary parts, exactly like combining like terms in a polynomial. Multiplication is FOIL, followed by replacing \(i^{2}\) with \(-1\).
Example 1: Adding and Subtracting
| Expression | Result |
|---|---|
| \((3+4i)+(2-5i)\) | \(5-i\) |
| \((7-2i)-(3+6i)\) | \(4-8i\) |
| \((-1+3i)+(4+i)\) | \(3+4i\) |
| \((5+2i)-(5-2i)\) | \(4i\) |
Example 2: Multiplying Complex Numbers
| Expression | Rule Applied | Result |
|---|---|---|
| \((2+3i)(1+4i)\) | FOIL, then \(i^{2}=-1\) | \(-10+11i\) |
| \((3-i)(2+i)\) | FOIL, then \(i^{2}=-1\) | \(7+i\) |
| \((4+i)(4-i)\) | Conjugate Product | \(17\) |
| \(i(3+2i)\) | Distribute, then \(i^{2}=-1\) | \(-2+3i\) |
Dividing Complex Numbers
A denominator with an \(i\) in it clears the same way a radical denominator does: multiply top and bottom by the denominator’s conjugate. The Conjugate Product rule guarantees the denominator becomes a real number with no \(i\) left at all.
Example 3: Dividing Complex Numbers
| Expression | Multiply By | Result |
|---|---|---|
| \(\dfrac{1}{1+i}\) | \(\dfrac{1-i}{1-i}\) | \(\dfrac{1}{2}-\dfrac{1}{2}i\) |
| \(\dfrac{3+i}{2-i}\) | \(\dfrac{2+i}{2+i}\) | \(1+i\) |
| \(\dfrac{2}{3+4i}\) | \(\dfrac{3-4i}{3-4i}\) | \(\dfrac{6}{25}-\dfrac{8}{25}i\) |
| \(\dfrac{4-2i}{1+i}\) | \(\dfrac{1-i}{1-i}\) | \(1-3i\) |
Worked Example A: Multiplying Two Complex Numbers
Multiply \((5+2i)(3-4i)\).
Step 1 — apply FOIL:
$$ (5+2i)(3-4i) = 15-20i+6i-8i^{2} $$
Step 2 — replace \(i^{2}\) with \(-1\):
$$ -8i^{2} = -8(-1) = 8 $$
Step 3 — combine real and imaginary parts:
$$ (15+8) + (-20+6)i = 23-14i $$
Worked Example B: Dividing with a Conjugate
Divide \(\dfrac{7+i}{3-2i}\).
Step 1 — multiply by the conjugate of the denominator:
$$ \frac{7+i}{3-2i} \cdot \frac{3+2i}{3+2i} $$
Step 2 — the denominator becomes a real number (Conjugate Product):
$$ (3-2i)(3+2i) = 3^{2}+2^{2} = 13 $$
Step 3 — expand the numerator:
$$ (7+i)(3+2i) = 21+14i+3i+2i^{2} = 21+17i-2 = 19+17i $$
Final answer: \(\dfrac{19+17i}{13} = \dfrac{19}{13}+\dfrac{17}{13}i\).
Worked Example C: A High Power of i
Simplify \(i^{23}\).
Step 1 — divide the exponent by 4 and find the remainder:
$$ 23 = 4 \cdot 5 + 3 $$
Step 2 — a remainder of 3 corresponds to \(i^{3}\):
$$ i^{23} = \left(i^{4}\right)^{5} \cdot i^{3} = 1^{5} \cdot i^{3} = i^{3} $$
Step 3 — evaluate:
$$ i^{3} = -i $$
Common Mistakes to Avoid
- Treating \(i^{2}\) as \(+1\) instead of \(-1\). This single substitution is the source of almost every step in this lesson — get it backward and everything downstream breaks.
- Multiplying straight across instead of using FOIL. \((a+bi)(c+di)\) needs all four products, exactly like multiplying two binomials.
- Forgetting to clear \(i\) from a denominator. A “final” answer with \(i\) still in the denominator isn’t fully simplified.
- Getting a conjugate’s sign backward. The conjugate of \(a-bi\) is \(a+bi\) — only the middle sign flips.
- Losing track of the power-of-i cycle. A remainder of \(0\) means \(1\), not \(i^{4}\) left unsimplified — always reduce to the final value.
Practice Problems
Work each problem yourself before opening the answer.
Problem 1. Add \((6+3i)+(2-7i)\).
Show answer
$$ (6+3i)+(2-7i) = (6+2)+(3-7)i = 8-4i $$
Answer: \(8-4i\)
Problem 2. Subtract \((4-5i)-(1+2i)\).
Show answer
$$ (4-5i)-(1+2i) = (4-1)+(-5-2)i = 3-7i $$
Answer: \(3-7i\)
Problem 3. Multiply \((2+i)(3+2i)\).
Show answer
$$ (2+i)(3+2i) = 6+4i+3i+2i^{2} = 6+7i-2 = 4+7i $$
Answer: \(4+7i\)
Problem 4. Multiply \((5-2i)(5+2i)\).
Show answer
$$ (5-2i)(5+2i) = 5^{2}+2^{2} = 25+4 = 29 $$
Answer: \(29\)
Problem 5. Simplify \(i^{10}\).
Show answer
$$ 10 = 4 \cdot 2 + 2, \qquad \text{remainder } 2 \to i^{2} = -1 $$
Answer: \(-1\)
Problem 6. Simplify \(i^{15}\).
Show answer
$$ 15 = 4 \cdot 3 + 3, \qquad \text{remainder } 3 \to i^{3} = -i $$
Answer: \(-i\)
Problem 7. Divide \(\dfrac{4}{1-i}\).
Show answer
$$ \frac{4}{1-i} \cdot \frac{1+i}{1+i} = \frac{4(1+i)}{1^{2}+1^{2}} = \frac{4+4i}{2} = 2+2i $$
Answer: \(2+2i\)
Problem 8. Divide \(\dfrac{2+3i}{1+i}\).
Show answer
Step 1 — multiply by the conjugate:
$$ \frac{2+3i}{1+i} \cdot \frac{1-i}{1-i} $$
Step 2 — denominator:
$$ 1^{2}+1^{2} = 2 $$
Step 3 — numerator:
$$ (2+3i)(1-i) = 2-2i+3i-3i^{2} = 2+i+3 = 5+i $$
Answer: \(\dfrac{5+i}{2} = \dfrac{5}{2}+\dfrac{1}{2}i\)
Problem 9. Multiply \(i(4-3i)\).
Show answer
$$ i(4-3i) = 4i-3i^{2} = 4i+3 = 3+4i $$
Answer: \(3+4i\)
Problem 10. Simplify \((3+i)^{2}\).
Show answer
$$ (3+i)^{2} = 9+6i+i^{2} = 9+6i-1 = 8+6i $$
Answer: \(8+6i\)
Quick Reference
| Situation | Move |
|---|---|
| \(i^{2}\) | Always replace with \(-1\) |
| Adding/subtracting | Combine real parts together, imaginary parts together |
| Multiplying | FOIL, then simplify \(i^{2}\) to \(-1\) |
| Dividing | Multiply by the conjugate of the denominator |
| \((a+bi)(a-bi)\) | Always equals \(a^{2}+b^{2}\) — a real number, no \(i\) left |
| \(i\) raised to a power | Divide the exponent by 4; the remainder gives \(1\), \(i\), \(-1\), or \(-i\) |
Once these rules feel automatic, they carry forward directly into solving any quadratic equation, even one with no real solutions, and into graphing complex numbers on the complex plane. Build speed with the Complex Number Generator, browse the rest of the Algebra lessons as they publish, revisit Radicals if the conjugate technique above felt unfamiliar, or explore the site’s calculators to check your own arithmetic against a worked result.