# Complex numbers and the imaginary unit i Canonical: https://duckyhelper.com/learn/algebra-2/complex-numbers/ Updated: 2026-10-01 A complex number has a real part and an imaginary part: \(a + bi\), where \(i = \sqrt{-1}\), so \(i^2 = -1\). Add and subtract by combining real parts with real parts and imaginary parts with imaginary parts. Multiply like binomials, then replace \(i^2\) with \(-1\). To divide, multiply the top and bottom by the conjugate of the bottom, which turns the denominator into a real number. ## The key idea No real number squared gives a negative, so mathematicians named a new number \(i\) with \(i^2 = -1\). Every complex number is a real part plus a real multiple of \(i\). Real numbers are complex numbers too: they just have \(b = 0\). $$ i = \sqrt{-1}, \qquad i^2 = -1, \qquad \sqrt{-k} = i\sqrt{k} \ \ (k > 0) $$ Powers of \(i\) repeat every 4 steps, because \(i^4 = (i^2)^2 = 1\): **The cycle of powers of i** | Power | \(i^1\) | \(i^2\) | \(i^3\) | \(i^4\) | \(i^5\) | | --- | --- | --- | --- | --- | --- | | Value | \(i\) | \(-1\) | \(-i\) | \(1\) | \(i\) (the cycle restarts) | The **conjugate** of \(a + bi\) is \(a - bi\). Their product is always a real number, which is why it is the tool for division: $$ (a + bi)(a - bi) = a^2 - b^2i^2 = a^2 + b^2 $$ ## Worked examples **Example 1: subtracting complex numbers** Problem: Simplify \((5 - 2i) - (3 - 7i)\). 1. The minus sign applies to both parts of the second number. $$ (5 - 2i) - (3 - 7i) = 5 - 2i - 3 + 7i $$ 2. Group real parts and imaginary parts. $$ 5 - 2i - 3 + 7i = (5 - 3) + (-2 + 7)i $$ 3. Combine. $$ (5 - 3) + (-2 + 7)i = 2 + 5i $$ Answer: \(2 + 5i\) **Example 2: multiplying** Problem: Simplify \((3 + 4i)(2 - i)\). 1. Multiply every term by every term, like FOIL. $$ (3 + 4i)(2 - i) = 6 - 3i + 8i - 4i^2 $$ 2. Replace \(i^2\) with \(-1\). Then \(-4i^2\) becomes \(+4\). $$ 6 - 3i + 8i - 4i^2 = 6 + 5i + 4 $$ 3. Combine the real parts. $$ 6 + 5i + 4 = 10 + 5i $$ Answer: \(10 + 5i\) **Example 3: dividing with the conjugate** Problem: Write \(\frac{7 + i}{1 - 2i}\) in the form \(a + bi\). 1. The conjugate of the bottom, \(1 - 2i\), is \(1 + 2i\). Multiply top and bottom by it. This does not change the value, because you are multiplying by 1. $$ \frac{7 + i}{1 - 2i} = \frac{(7 + i)(1 + 2i)}{(1 - 2i)(1 + 2i)} $$ 2. Top: \(7 + 14i + i + 2i^2 = 7 + 15i - 2\). $$ (7 + i)(1 + 2i) = 5 + 15i $$ 3. Bottom: \(a^2 + b^2\) with \(a = 1\), \(b = 2\). $$ (1 - 2i)(1 + 2i) = 5 $$ 4. Divide each part by 5. $$ \frac{5 + 15i}{5} = 1 + 3i $$ Answer: \(1 + 3i\) **Example 4 (test-hard): a quadratic with complex roots** Problem: Solve \(x^2 - 6x + 13 = 0\). 1. Use the quadratic formula with \(a = 1\), \(b = -6\), \(c = 13\). Start with the discriminant. $$ b^2 - 4ac = 36 - 52 = -16 $$ 2. The discriminant is negative, so the roots are not real. Its square root is \(\sqrt{-16} = 4i\). $$ x = \frac{6 \pm 4i}{2} $$ 3. Divide both parts by 2. $$ x = 3 \pm 2i $$ 4. Check \(x = 3 + 2i\): \((3 + 2i)^2 = 9 + 12i - 4 = 5 + 12i\), and \(5 + 12i - 6(3 + 2i) + 13 = 5 + 12i - 18 - 12i + 13 = 0\). Answer: \(x = 3 + 2i\) or \(x = 3 - 2i\) ## Common mistakes - **Leaving \(i^2\) in the answer.** \(i^2\) is just \(-1\). Fix: after multiplying, replace every \(i^2\) and combine the real parts. - **Dropping the sign on the second number when subtracting.** \(-(3 - 7i)\) is \(-3 + 7i\). Fix: distribute the minus to both parts. - **Using \(a^2 - b^2\) for the conjugate product.** \((1 - 2i)(1 + 2i)\) is \(1 + 4 = 5\), not \(1 - 4\). Fix: the \(i^2\) turns the minus into a plus, so it is \(a^2 + b^2\). - **Multiplying square roots of negatives directly.** \(\sqrt{-4} \cdot \sqrt{-9}\) is \(2i \cdot 3i = -6\), not \(\sqrt{36} = 6\). Fix: rewrite each \(\sqrt{-k}\) as \(i\sqrt{k}\) first. - **Dividing only the real part by the denominator.** \(\frac{5 + 15i}{5}\) is \(1 + 3i\), not \(1 + 15i\). Fix: divide both parts. ## Quick methods > **Tip: Big powers of i: use the remainder** > > Divide the exponent by 4 and look at the remainder: 0 gives 1, 1 gives \(i\), 2 gives \(-1\), 3 gives \(-i\). For \(i^{43}\): \(43 = 4 \cdot 10 + 3\), so \(i^{43} = -i\). This always works. > **Tip: Conjugate products in one step** > > \((a + bi)(a - bi) = a^2 + b^2\). So \((4 + 3i)(4 - 3i) = 16 + 9 = 25\) with no FOIL at all. ## Practice **5 practice questions** 1. Simplify \(i^{27}\). A. \(i\) B. \(-i\) C. \(1\) D. \(-1\) Answer: \(-i\). \(27 = 4 \cdot 6 + 3\), so \(i^{27} = i^3 = -i\). Picking \(i\) or \(-1\) usually means the remainder was counted wrong. 2. Simplify \((4 - 3i)(4 + 3i)\). A. \(7\) B. \(25\) C. \(16 + 9i\) D. \(7 + 24i\) Answer: \(25\). Conjugates multiply to \(a^2 + b^2 = 16 + 9 = 25\). The answer 7 comes from treating \(-9i^2\) as \(-9\). \(7 + 24i\) is \((4 + 3i)^2\), a different product. 3. Simplify \((2 + 5i) + (-6 + i) - (1 - 3i)\). A. \(-5 + 9i\) B. \(-5 + 3i\) C. \(-3 + 9i\) D. \(-5 - 9i\) Answer: \(-5 + 9i\). Real parts: \(2 - 6 - 1 = -5\). Imaginary parts: \(5 + 1 + 3 = 9\). \(-5 + 3i\) forgets that subtracting \(-3i\) adds \(3i\). 4. Write \(\frac{10}{3 + i}\) in the form \(a + bi\). A. \(3 - i\) B. \(3 + i\) C. \(\frac{10}{3} + 10i\) D. \(\frac{30 - 10i}{8}\) Answer: \(3 - i\). Multiply top and bottom by \(3 - i\): \(\frac{10(3 - i)}{9 + 1} = \frac{30 - 10i}{10} = 3 - i\). The choice with 8 on the bottom used \(9 - 1\) instead of \(9 + 1\). 5. What are the solutions of \(x^2 + 4x + 20 = 0\)? A. \(-2 \pm 4i\) B. \(2 \pm 4i\) C. \(-2 \pm 8i\) D. \(-4 \pm 8i\) Answer: \(-2 \pm 4i\). The discriminant is \(16 - 80 = -64\), and \(\sqrt{-64} = 8i\). So \(x = \frac{-4 \pm 8i}{2} = -2 \pm 4i\). \(-4 \pm 8i\) forgets to divide by 2, and \(2 \pm 4i\) drops the minus on b. ## Frequently asked questions ### What is i in math? \(i\) is the imaginary unit, the number whose square is \(-1\). It lets you take square roots of negative numbers: \(\sqrt{-25} = 5i\). It follows all the usual rules of algebra, with one extra fact: whenever you see \(i^2\), replace it with \(-1\). ### Why do complex roots come in pairs? When a quadratic has real coefficients, the quadratic formula gives \(\frac{-b \pm \sqrt{D}}{2a}\). If D is negative, the plus and minus give \(p + qi\) and \(p - qi\): a conjugate pair. So if \(3 + 2i\) is a root, \(3 - 2i\) is a root too. ### Is a real number a complex number? Yes. A real number like 7 is \(7 + 0i\), a complex number whose imaginary part is zero. Complex numbers include all real numbers, all imaginary numbers like \(4i\), and mixes like \(2 - 3i\). ### Why do we multiply by the conjugate when dividing? A complex number in the denominator is not in standard \(a + bi\) form. Multiplying top and bottom by the conjugate makes the bottom \(a^2 + b^2\), a plain real number, and does not change the value. Then you can split the fraction into a real part and an imaginary part. ## Related - [Completing the square](https://duckyhelper.com/learn/algebra-2/completing-the-square/) - [How to solve quadratic equations](https://duckyhelper.com/learn/algebra-1/solving-quadratics/) - [How to simplify square roots and radicals](https://duckyhelper.com/learn/algebra-1/simplifying-radicals/) - [Quadratics on the SAT](https://duckyhelper.com/learn/sat-math/quadratics/) - [Algebra 2 study guides](https://duckyhelper.com/learn/algebra-2/) ## Try asking Ducky - "Why do I multiply by the conjugate and not just the bottom?" - "I got 10 - 5i for the multiplication one. Can you check my signs?" - "Quiz me on powers of i until I get five right in a row." ## Get DuckyHelper Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. 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