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⚔️ Algebra Kingdom · Algebra

Complex Numbers

Extend the number system with i = sqrt(-1): simplify powers of i, add, subtract, multiply and divide complex numbers, and solve the quadratics whose discriminant is negative.

In short

  • i is defined by i2 = -1, which makes sqrt(-1) = i and lets every square root of a negative number be written down.
  • Every complex number is a + bi with a and b real: a is the real part, b is the number in front of i, and an ordinary number is just a + 0i.
  • The powers of i cycle through i, -1, -i, 1, so in depends only on the remainder when n is divided by 4.
  • Adding and subtracting is collecting like terms; multiplying is expanding brackets and then replacing i2 with -1, which is the only new rule in the topic.
  • The conjugate of a + bi is a - bi, and a number times its conjugate is the real number a2 + b2 — which is what clears an i out of a denominator.
  • A negative discriminant no longer means there is no answer: it means the two solutions are the conjugate pair p + qi and p - qi.
  • The modulus |a + bi| = sqrt(a^2 + b2) is the distance from 0 to the point (a, b), so it is never negative.

The number the line had no room for

Herald Vex keeps one question chalked above the door of Quadratic Castle: what squares to give -1?

Nothing on the number line does. A positive number squared is positive, a negative number squared is positive, and 0 squared is 0. So sqrt(-1) is not anywhere on that line — and for a long time that was the end of the discussion.

The Castle's answer is to widen the world rather than close the question. Define a brand new number, called i, by the one property that was missing:

i2 = -1, so sqrt(-1) = i

Everything else in this topic is a consequence of that single line. Once i exists, every square root of a negative number can be written down:

sqrt(-25) = sqrt(25) * sqrt(-1) = 5i sqrt(-50) = sqrt(50) * sqrt(-1) = 5 sqrt(2) i

A number of the form a + bi, with a and b real, is called a complex number. The a is its real part and the b is its imaginary part — and note that b is the *number in front of* i, not the whole term. In 3 - 7i the real part is 3 and the imaginary part is -7.

Ordinary numbers have not gone anywhere. 6 is 6 + 0i, and 4i is 0 + 4i. The real numbers are simply the complex numbers whose imaginary part happens to be 0.

One warning belongs here, at the start, because it is the trap that catches everyone. The rule sqrt(m) * sqrt(n) = sqrt(mn) is only true when both radicands are positive. So

sqrt(-9) * sqrt(-4) = 3i * 2i = 6 i2 = -6

and not sqrt(36) = 6. Rewrite each root with an i before you multiply anything.

Powers of i go round in a circle of four

Start multiplying i by itself and something pleasant happens: the answers repeat.

i1 = i i2 = -1 i3 = i2 * i = -i i4 = i2 * i2 = (-1)(-1) = 1 i5 = i4 * i = i ... and round it goes again

Every fourth power lands back on 1, so the powers of i cycle through i, -1, -i, 1 for ever. To simplify in, all you need is the remainder when n is divided by 4:

  • remainder 0 gives 1
  • remainder 1 gives i
  • remainder 2 gives -1
  • remainder 3 gives -i

So for i47: 47 = 4 * 11 + 3, the remainder is 3, and i47 = -i. You never have to multiply forty-seven things together.

Two details are worth saying out loud. The cycle starts at i1 = i and closes at i4 = 1 — counting from i0 shifts every answer one step round the circle. And half the cycle is negative, so check whether the remainder is 2 or 3 before you drop a minus sign.

The same idea handles a power of a whole imaginary term, because a power of a product is a power of each factor:

(2i)3 = 23 * i3 = 8 * (-i) = -8i

Arithmetic: one new rule and no others

Adding and subtracting complex numbers is collecting like terms. Real parts join real parts, i parts join i parts, and nothing crosses between them — a 3 and a 4i can no more be added than 3 apples and 4 arrows.

(5 + 2i) + (1 - 6i) = (5 + 1) + (2 - 6)i = 6 - 4i

A subtracted bracket needs the usual care: the minus reaches both numbers inside it.

(5 + 2i) - (3 + 4i) = (5 - 3) + (2 - 4)i = 2 - 2i

Multiplying is expanding brackets, exactly as in any algebra, plus the one replacement i2 = -1 at the end. Every term in the first bracket meets every term in the second:

(3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10 i2 = 12 - 7i - 10(-1) = 22 - 7i

Watch the last step. Because i2 = -1, the term -10 i2 turns into +10 — a minus that becomes a plus. Getting that sign backwards is the single most common error in the whole topic.

Squaring works the same way, and a bracket squared is never each term squared:

(3 + 4i)2 = 9 + 24i + 16 i2 = 9 + 24i - 16 = -7 + 24i

One product is special. The conjugate of a + bi is a - bi: same real part, opposite imaginary part. Multiply a complex number by its own conjugate and the two middle terms cancel:

(3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16 i2 = 9 + 16 = 25

The result is always a2 + b2 — real, positive, and no i in sight. Its square root, sqrt(a^2 + b2), is called the modulus |a + bi|: plot a + bi as the point (a, b) and the modulus is its distance from 0. For 3 + 4i that distance is sqrt(9 + 16) = 5, the 3-4-5 triangle in disguise.

Division: send the conjugate downstairs

A complex number is never left in a denominator, for the same reason a surd is never left there: you cannot see how big it is.

The fix is the conjugate. Multiply the top and the bottom by the conjugate of the denominator — that is multiplying by 1, so the value does not change — and the bottom turns real.

(11 + 2i) / (3 - 4i)

Conjugate of the bottom: 3 + 4i. Multiply both parts by it.

Bottom: (3 - 4i)(3 + 4i) = 32 + 42 = 25 Top: (11 + 2i)(3 + 4i) = 33 + 44i + 6i + 8 i2 = 33 + 50i - 8 = 25 + 50i

(25 + 50i) / 25 = 1 + 2i

Three things to hold on to.

  • The conjugate flips the sign of the i term. Multiplying by the denominator itself leaves an i downstairs and gets you nowhere.
  • The new denominator is c2 + d2, an addition, because i2 = -1 turns the subtraction in the difference of squares into a plus.
  • Both parts of the numerator get divided by that denominator. Sometimes they divide exactly; sometimes they do not, and the answer is written with fractions, as in (4 + 7i)/5 = 4/5 + (7/5)i.

Quadratics that used to have no answer — and how to type one

You already know that the discriminant b2 - 4ac counts the real solutions of ax2 + bx + c = 0: positive gives two, zero gives one, negative gives none. That last case is no longer the end of the road. A negative discriminant means the square root in the quadratic formula is a square root of a negative number, so both solutions are complex.

Solve x2 - 6x + 13 = 0. Here a = 1, b = -6, c = 13.

b2 - 4ac = 36 - 52 = -16 sqrt(-16) = 4i x = (6 +- 4i) / 2 = 3 +- 2i

The two solutions are 3 + 2i and 3 - 2i — always a conjugate pair, whenever a, b and c are real. Check the divide: everything in the numerator is divided by 2a, not just the first part. And a simpler equation such as (x - 3)2 = -16 needs no formula at all: take the square root of both sides to get x - 3 = +- 4i, so x = 3 +- 4i.

Typing your answer here. A complex number a + bi is typed as the pair (a, b), and every prompt in this skill says so in the same breath.

  • 3 + 2i is typed (3, 2); 3 - 2i is typed (3, -2).
  • A real answer is (a, 0): -6 is typed (-6, 0).
  • A pure imaginary answer is (0, b): 4i is typed (0, 4).
  • Where a part is a fraction, type it as a fraction in lowest terms: 1/2 + (3/2)i is (1/2, 3/2).
  • A quadratic has two solutions and a pair cannot hold both, so the prompt asks for the one whose imaginary part is positive — for 3 +- 2i, that is (3, 2).

The other answer shapes are the ones a pair cannot carry. A power of i is chosen from 1, i, -1, -i. A root that does not tidy into whole numbers, such as sqrt(-50) = 5 sqrt(2) i, is chosen from a list. And a modulus is a plain whole number, so |5 + 12i| is typed 13.

Worked examples

Example 1

Simplify i103.

  1. Only the remainder on division by 4 matters, because i4 = 1 and every complete block of four powers multiplies to 1.
  2. Divide: 103 = 4 * 25 + 3, so the remainder is 3.
  3. i103 = (i4)25 * i3 = 1 * i3 = i3.
  4. i3 = i2 * i = (-1) * i = -i.
  5. So i103 = -i.

Example 2

Expand (4 - 3i)(2 + 5i) and write the result in the form a + bi.

  1. Every term in the first bracket meets every term in the second: 4 * 2, 4 * 5i, -3i * 2 and -3i * 5i.
  2. That gives 8 + 20i - 6i - 15 i2.
  3. Replace i2 with -1: the last term -15 i2 becomes +15.
  4. Collect the real parts: 8 + 15 = 23. Collect the i parts: 20i - 6i = 14i.
  5. The product is 23 + 14i, typed as the pair (23, 14).

Example 3

Simplify (7 + i) / (2 - 3i), and write the result in the form a + bi.

  1. The conjugate of the denominator 2 - 3i is 2 + 3i. Multiply the top and the bottom by it.
  2. Bottom: (2 - 3i)(2 + 3i) = 22 + 32 = 4 + 9 = 13, a real number.
  3. Top: (7 + i)(2 + 3i) = 14 + 21i + 2i + 3 i2 = 14 + 23i - 3 = 11 + 23i.
  4. Divide both parts by 13: 11/13 + (23/13)i.
  5. Neither fraction cancels, so the answer is 11/13 + (23/13)i, typed as the pair (11/13, 23/13).

Example 4

Solve 2x2 - 4x + 5 = 0, and give the solution whose imaginary part is positive.

  1. Read off a = 2, b = -4, c = 5, then work out the discriminant before anything else.
  2. b2 - 4ac = 16 - 40 = -24 — negative, so both solutions are complex.
  3. sqrt(-24) = sqrt(24) * i = 2 sqrt(6) i.
  4. x = (4 +- 2 sqrt(6) i) / 4 = 1 +- (sqrt(6)/2) i, so the two solutions are a conjugate pair as expected.
  5. The one with the positive imaginary part is 1 + (sqrt(6)/2) i.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Write sqrt(-49) in the form a + bi. Give your answer as the pair (a, b), meaning a + bi.

Answer: (0, 7)

  1. sqrt(-49) = sqrt(49) * sqrt(-1)
  2. sqrt(49) = 7 and sqrt(-1) = i
  3. sqrt(-49) = 7i, which is 0 + 7i in the form a + bi.
  4. Pair: (0, 7)

Problem 2

Difficulty 3 of 5

Simplify i107.

  1. 1
  2. -1
  3. i
  4. -i

Answer: D. -i

  1. Divide the exponent by 4: 107 = 4 * 26 + 3
  2. i4 = 1, so every complete block of four powers multiplies to 1.
  3. i107 = i3 = -i

Problem 3

Difficulty 4 of 5

Simplify (-11 + 8i) - (-3 + 4i) + (1 - 8i), and write the result in the form a + bi. Give your answer as the pair (a, b), meaning a + bi.

Answer: (-7, -4)

  1. Drop the brackets, changing signs where a bracket is subtracted.
  2. Real parts: -11 - (-3) + 1 = -7
  3. Imaginary parts: 8 - 4 + (-8) = -4
  4. Result: -7 - 4i
  5. Pair: (-7, -4)

Common mistakes

  • Multiplying the radicands first: writing sqrt(-9) * sqrt(-4) as sqrt(36) = 6. The rule sqrt(m) * sqrt(n) = sqrt(mn) needs positive radicands, and the honest answer is 3i * 2i = 6 i2 = -6.
  • Adding instead of subtracting when i2 appears: leaving -10 i2 as -10 rather than +10, which flips the sign of the real part of every product.
  • Letting a subtracted bracket reach only the first number: writing (5 + 2i) - (3 + 4i) as 2 + 6i instead of 2 - 2i.
  • Squaring each term of a bracket: writing (3 + 4i)2 as 9 - 16 and forgetting the middle term 24i.
  • Multiplying top and bottom by the denominator instead of its conjugate, which leaves an i downstairs and solves nothing — the sign of the i term has to flip.
  • Dividing only the first part of a numerator by 2a in the quadratic formula, so 3 +- 2i comes out as 6 +- 2i.
  • Giving the solution with the negative imaginary part when the question asked for the positive one, or typing the pair the wrong way round: a + bi is (a, b), real part first.
  • Stopping one step early on a modulus and reporting a2 + b2, or adding the parts instead of squaring them: |5 + 12i| is 13, not 169 and not 17.

What you should be able to do

  • Write the square root of a negative number in terms of i, and simplify a power of i.
  • Add, subtract and multiply complex numbers, and write the result in the form a + bi.
  • Divide complex numbers by multiplying by the conjugate of the denominator.
  • Solve a quadratic equation with a negative discriminant and write its complex solutions.

Where this fits in the curriculum

Common Core

  • HSN-CN.A.1

    High school — Know there is a complex number i such that i² = −1, and every complex number has the form a + bi with a and b real.

  • HSN-CN.A.2

    High school — Use the relation i² = −1 and the commutative, associative and distributive properties to add, subtract and multiply complex numbers.

  • HSN-CN.C.7

    High school — Solve quadratic equations with real coefficients that have complex solutions.

  • HSA-REI.B.4.B

    High school — Solve quadratic equations by inspection, taking square roots, completing the square, the quadratic formula and factoring, and recognise when the quadratic formula gives complex solutions and write them as a ± bi.

  • HSN-CN.A.3

    High school — Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers.

    HSN-CN.A.3 is a (+) standard — beyond the college- and career-ready threshold, i.e. precalculus rather than Algebra II.

SAT

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