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Complex Numbers

What happens when you take the square root of a negative number, how to do arithmetic with the answer, and why every quadratic equation has solutions.

No real number squared gives −1, so x2 + 1 = 0 has no real solution. Mathematicians fixed that by naming a new number, i, with i2 = −1. Numbers like 3 + 2i are called complex numbers. You can add, multiply and divide them, and you can draw them as points on a plane.

Start: the number iJump to multiplyingJump to the self-guided quiz
The complex plane: 3 + 2i and i times itRealImaginary−4−3−2−112345−2−11234503 + 2i−2 + 3i
The point 3 + 2i sits 3 right and 2 up. Multiplying it by i gives i(3 + 2i) = −2 + 3i: the arrow turns a quarter turn counterclockwise around 0.

The number i

The imaginary unit i is defined by i2 = −1, and we write √−1 = i. For any positive number n, √−n = i√n. So a square root of a negative number is a real square root with an i in front. The name "imaginary" is just a name: these numbers are as useful as any others, in fields like electrical engineering.

Example 1: Square roots of negative numbers

√−25 = √−1 · √25Split off the −1√−25 = 5i√25 = 5 and √−1 = i√−49 = 7iThe same way

Example 2: A root that isn't a whole number

√−12 = i√12√−12 = 2i√3√12 = √4 · √3 = 2√3

Example 3: Write roots with i before multiplying

√−4 · √−9 = 2i · 3iChange each root to i form first√−4 · √−9 = 6i2√−4 · √−9 = −6i2 = −1. Multiplying under one root first would give √36 = 6, which has the wrong sign

Powers of i

Multiply by i over and over and the answers repeat in a cycle of four: i1 = i, i2 = −1, i3 = i2 · i = −i, and i4 = (i2)2 = 1. Then i5 = i and the cycle starts again. To simplify a big power, divide the exponent by 4 and use the remainder.

PowerValue
i^1i
i^2−1
i^3−i
i^41
i^5i

Example 4: Simplify a large power

i23 = i20 · i323 divided by 4 is 5 remainder 3i23 = 1 · i3i20 = (i4)5 = 1i23 = −i

Example 5: Another one

i42 = i40 · i242 divided by 4 is 10 remainder 2i42 = −1

The form a + bi

A complex number is written a + bi, where a and b are real numbers. The number a is the real part and b is the imaginary part. If b = 0 the number is real, so every real number is also a complex number. If a = 0 and b is not 0, the number, like 4i, is called pure imaginary. Two complex numbers are equal only when their real parts match and their imaginary parts match.

Example 6: Name the parts

−7 + 0.5iReal part −7, imaginary part 0.56Real part 6, imaginary part 0−3iReal part 0, imaginary part −3

Example 7: Use equality to find unknowns

2x + 6i = 8 + 3yiMatch the parts2x = 8 and 6 = 3yReal parts equal, imaginary parts equalx = 4 and y = 2

Adding and subtracting

Add or subtract complex numbers the way you combine like terms: real parts with real parts, imaginary parts with imaginary parts. When you subtract, the minus sign changes the sign of both parts of the second number.

Example 8: Add

(4 + 3i) + (2 − 5i) = (4 + 2) + (3 − 5)iGroup the parts(4 + 3i) + (2 − 5i) = 6 − 2i

Example 9: Subtract

(1 − 2i) − (−3 + 4i) = 1 − 2i + 3 − 4iThe minus changes both signs(1 − 2i) − (−3 + 4i) = 4 − 6i

Multiplying

Multiply complex numbers like binomials, using FOIL or the distributive property. Every i2 that appears becomes −1, which turns that term into a real number. Then combine the real parts and the imaginary parts.

Example 10: Multiply by a real number and by i

3i(2 − 5i) = 6i − 15i2Distribute3i(2 − 5i) = 6i + 15−15i2 = −15(−1) = 153i(2 − 5i) = 15 + 6iWrite the real part first

Example 11: FOIL two complex numbers

(2 + 3i)(4 − i) = 8 − 2i + 12i − 3i2FOIL(2 + 3i)(4 − i) = 8 + 10i + 3−3i2 = 3(2 + 3i)(4 − i) = 11 + 10i

Example 12: Square a complex number

(1 + 2i)2 = 1 + 4i + 4i2(a + b)2 = a2 + 2ab + b2(1 + 2i)2 = −3 + 4i4i2 = −4

Conjugates and dividing

The conjugate of a + bi is a − bi: same real part, opposite imaginary part. A number times its conjugate is always a real number, because the i terms cancel: (a + bi)(a − bi) = a2 + b2. That is the key to dividing. To divide, multiply the top and the bottom by the conjugate of the bottom. The bottom becomes real, and you can split the answer into a + bi form.

Example 13: A number times its conjugate

(4 − 3i)(4 + 3i) = 16 + 12i − 12i − 9i2FOIL(4 − 3i)(4 + 3i) = 16 + 9The middle terms cancel, and −9i2 = 9(4 − 3i)(4 + 3i) = 25A real number

Example 14: Divide by a complex number

5 + 10i1 + 2i = 5(1 + 2i)1 + 2iHere the top factors, so this one is quick5 + 10i1 + 2i = 5

Example 15: Divide using the conjugate

4 + 7i2 + i = (4 + 7i)(2 − i)(2 + i)(2 − i)Multiply top and bottom by the conjugate 2 − i4 + 7i2 + i = 8 − 4i + 14i − 7i24 + 1FOIL the top; the bottom is 22 + 124 + 7i2 + i = 15 + 10i5−7i2 = 74 + 7i2 + i = 3 + 2iDivide each part by 5. Check: (2 + i)(3 + 2i) = 6 + 4i + 3i + 2i2 = 4 + 7i ✓

The complex plane and absolute value

A real number is a point on a number line. A complex number needs two directions, so we draw it on the complex plane: the horizontal axis is the real axis and the vertical axis is the imaginary axis. The number a + bi is the point (a, b). Its conjugate is its mirror image across the real axis.

The absolute value |a + bi| is the distance from the point to 0. By the Pythagorean theorem, |a + bi| = √a2 + b2.

Points in the complex plane and the absolute value of 3 + 4iRealImaginary−5−3−2−112345−5−4−3−11234503 + 4i3 − 4i−4−2i
The point 3 + 4i is at (3, 4), and its conjugate 3 − 4i is at (3, −4). The real number −4 sits on the real axis, and −2i sits on the imaginary axis. The teal segment from 0 to 3 + 4i has length √32 + 42 = 5.

Example 16: Find an absolute value

|−5 + 12i| = √(−5)2 + 122|−5 + 12i| = √16925 + 144 = 169|−5 + 12i| = 13

Quadratics with complex solutions

In the quadratic formula x = −b + √b2 − 4ac2a (and the same with a minus sign before the root), the discriminant b2 − 4ac decides what kind of solutions you get. When it is negative, the parabola never crosses the x-axis and there are no real solutions. With i, the square root of the negative discriminant still makes sense, so the equation has two complex solutions, and they are conjugates of each other.

Example 17: Solve by taking square roots

x2 + 9 = 0x2 = −9x = 3i or x = −3i√−9 = 3i. Check: (3i)2 + 9 = −9 + 9 = 0 ✓

Example 18: Use the quadratic formula

x2 + 2x + 5 = 0a = 1, b = 2, c = 5b2 − 4ac = 4 − 20 = −16Negative, so the solutions are not realx = −2 + 4i2 or x = −2 − 4i2√−16 = 4ix = −1 + 2i or x = −1 − 2iA conjugate pair. Check: (−1 + 2i)2 = −3 − 4i, so x2 + 2x + 5 becomes (−3 − 4i) + (−2 + 4i) + 5 = 0 ✓

Common mistakes

✗Writing i2 = 1.✓By definition i2 = −1. That one fact drives every calculation on this page.
✗Writing √−4 · √−9 = √36 = 6.✓The rule √a · √b = √ab only works when a and b are not negative. Write 2i · 3i = 6i2 = −6.
✗Forgetting to change both signs when subtracting.✓(1 − 2i) − (−3 + 4i) is 1 − 2i + 3 − 4i, not 1 − 2i − 3 + 4i.
✗Leaving i2 in an answer.✓Replace every i2 with −1 and combine, so the answer is in a + bi form.
✗Dividing only the real parts, or multiplying only the bottom by the conjugate.✓Multiply the top and the bottom by the conjugate, so you are multiplying by 1.

Practice problems

Try each one, then open the answer to check your work.

  1. Simplify √−64.
    Show answer
    √−64 = 8i
  2. Simplify i35.
    Show answer
    i35 = i32 · i335 divided by 4 is 8 remainder 3i35 = −i
  3. Add (7 − 2i) + (−3 + 6i).
    Show answer
    (7 − 2i) + (−3 + 6i) = 4 + 4i
  4. Subtract (5 + i) − (2 − 3i).
    Show answer
    (5 + i) − (2 − 3i) = 5 + i − 2 + 3i(5 + i) − (2 − 3i) = 3 + 4i
  5. Multiply (3 − 2i)(1 + 4i).
    Show answer
    (3 − 2i)(1 + 4i) = 3 + 12i − 2i − 8i2FOIL(3 − 2i)(1 + 4i) = 11 + 10i−8i2 = 8
  6. Multiply (6 + 5i)(6 − 5i).
    Show answer
    (6 + 5i)(6 − 5i) = 36 + 25Conjugates: a2 + b2(6 + 5i)(6 − 5i) = 61
  7. Divide 10 + 5i2 − i.
    Show answer
    10 + 5i2 − i = (10 + 5i)(2 + i)(2 − i)(2 + i)Multiply by the conjugate10 + 5i2 − i = 15 + 20i520 + 10i + 10i + 5i2 = 15 + 20i, and 4 + 1 = 510 + 5i2 − i = 3 + 4i
  8. Find |8 − 6i|.
    Show answer
    |8 − 6i| = √64 + 36|8 − 6i| = 10
  9. Solve x2 − 4x + 13 = 0.
    Show answer
    b2 − 4ac = 16 − 52 = −36x = 4 + 6i2 or x = 4 − 6i2√−36 = 6ix = 2 + 3i or x = 2 − 3i

History: numbers nobody wanted

Square roots of negative numbers first forced their way into mathematics through cubic equations, not quadratics. In 1545 the Italian doctor and mathematician Girolamo Cardano published Ars Magna (The Great Art), with methods for solving cubic and quartic equations. According to the MacTutor History of Mathematics archive, the cubic formula sometimes involved square roots of negative numbers even when the final answer was an ordinary number.

Rafael Bombelli took those strange roots seriously. The first three books of his Algebra were published in 1572, and MacTutor credits him as the first person to write down the rules for adding, subtracting and multiplying complex numbers.

The names came later still. MacTutor lists the symbol i for the square root of −1 among Leonhard Euler's notations, dated 1777, and says Carl Friedrich Gauss introduced the term ‘complex number’ in 1831.

Drawing complex numbers as points came much later. The Norwegian-Danish surveyor Caspar Wessel presented a geometric interpretation to the Royal Danish Academy in 1797, published in 1799. Jean-Robert Argand produced a similar one in 1806, and today the picture is often called an Argand diagram even though Wessel's work came first.

Connections

This unit picks up where Quadratic Equations left off: a negative discriminant now means two complex solutions instead of no solution. The rules for square roots come from Exponents and Radicals, and multiplying complex numbers uses the same FOIL steps as Polynomials and Factoring.

Dividing with a conjugate works like rationalizing a denominator, and the absolute value |a + bi| is the Pythagorean theorem on the complex plane. For more on that theorem, see The Pythagorean Theorem in Jugami Geometry.

Self-guided quiz

Pick a level, or Mixed for a bit of everything. Each question has one answer, a complex number in the form a + bi, like 3 − 2i, 5 or −4i. You can type z = or x = in front. Your score is just for you: nothing is sent anywhere.

Sources

  • J J O'Connor and E F Robertson, “Girolamo Cardano”, MacTutor History of Mathematics, University of St Andrews: Cardano biography (Ars Magna, 1545, and square roots of negative numbers in the cubic formula)
  • J J O'Connor and E F Robertson, “Rafael Bombelli”, MacTutor: Bombelli biography (Algebra, 1572, and the first written rules for complex arithmetic)
  • J J O'Connor and E F Robertson, “Leonhard Euler”, MacTutor: Euler biography (the symbol i for the square root of −1, 1777)
  • J J O'Connor and E F Robertson, “The fundamental theorem of algebra”, MacTutor: history topic (Gauss introduced the term ‘complex number’ in 1831)
  • J J O'Connor and E F Robertson, “Caspar Wessel”, MacTutor: Wessel biography (paper presented 1797, published 1799)
  • J J O'Connor and E F Robertson, “Jean-Robert Argand”, MacTutor: Argand biography (geometric interpretation, 1806)
  • Common Core State Standards for Mathematics: The Complex Number System (N-CN.A.1, N-CN.A.2, N-CN.C.7)

Credits: all diagrams on this page are original SVG drawings made for Jugami. Every answer was double-checked.