Complex Numbers — Print

One new number, i, and every polynomial equation finally has a solution.

What the practice questions cover

These questions use 6 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Find a number on the line whose square is -1.

Positive times positive is positive. Negative times negative is also positive. Zero squares to zero.

nowhere on the line does x² = −1 -3 -2 -1 0 1 2 3 square any of these and you land on the right half so step off the line i real imaginary one new number, and every equation has an answer

So no point anywhere on that line works. The line is not big enough for the question.

Mathematicians had two choices: declare the question meaningless, or step off the line. They stepped off.

Call the new direction i, with i2 = -1. The line becomes a plane, and every quadratic that had "no solution" turns out to have two.

A complex number is a point, not a place on the line real imaginary 3 + 2i 3 right 2 up i real part left-right, imaginary part up-down

Powers of i go round in fours

i1 = i, i2 = -1, i3 = -i, i4 = 1 — and then it starts again.

So to find i39, divide 39 by 4 and keep the remainder 3: the answer is i3 = -i.

The whole cycle is four steps long, which is why only the remainder matters.

Divide by four, keep the remainder

Remainder 0 → 1, 1 → i, 2 → -1, 3 → -i. Four cases, and they never change.

Adding and subtracting

Treat i like any other letter: real with real, imaginary with imaginary.

(3 + 4i) + (5 - 2i) = 8 + 2i. Nothing new is happening here.

The answer is always written a + bi, real part first.

It is collecting like terms

If you can simplify 3x + 5x, you can add complex numbers. The only difference is what happens when two is meet.

Multiplying, and the moment i2 appears

Multiply out as usual, then replace every i2 with -1. That replacement is the only new rule.

(2+3i)(1-4i) = 2 - 8i + 3i - 12i2, and -12i2 becomes +12.

Collecting up: 14 - 5i. Notice the real part grew, because the i2 term crossed over.

The sign flip is easy to miss

-12i2 = -12(-1) = +12. Two negatives make a positive here, and skipping that step is the commonest slip in the unit.

Dividing by using the conjugate

You cannot leave i in a denominator, so multiply top and bottom by the conjugate of the bottom.

(a+bi)(a-bi) = a2 + b2, a plain real number, because the middle terms cancel and -b2i2 turns positive.

That single move clears the denominator every time, with no cases to consider.

Why the conjugate

It is the difference of two squares, arranged so the leftover i2 becomes real. The same trick rationalises 12+√3.

Where complex roots come from

In the quadratic formula, a negative under the square root used to end the problem.

Now √-16 = 4i, so the formula keeps going and returns two complex roots.

They always arrive as a conjugate pair — if 3 + 2i is a root of a real polynomial, so is 3 - 2i.

What the graph shows

A negative discriminant means the parabola misses the x-axis. The roots are still there; they simply are not on the picture.

Worked examples

A high power of i

Simplify i47.

  1. Divide 47 by 4: 47 = 4(11) + 3. The powers repeat every four steps.
  2. The remainder is 3, so i47 = i3. Only the remainder survives.
  3. And i3 = i2 · i = -i. Because i2 = -1.

-i

Multiplying

Expand (2 + 3i)(1 - 4i).

  1. Multiply out: 2 - 8i + 3i - 12i2. Ordinary expansion, nothing special yet.
  2. Replace i2 with -1: -12i2 = +12. The one new rule, and the one easily missed.
  3. Collect: (2 + 12) + (-8 + 3)i = 14 - 5i. Real with real, imaginary with imaginary.

14 - 5i

Dividing

Simplify 52 - i.

  1. Multiply top and bottom by the conjugate 2 + i. This is what clears i from the denominator.
  2. Bottom: (2-i)(2+i) = 4 - i2 = 4 + 1 = 5. The middle terms cancel, leaving a real number.
  3. Top: 5(2+i) = 10 + 5i, so the answer is 10+5i5 = 2 + i. Divide both parts by 5.

2 + i

Words to know

Imaginary unit
i, defined by i2 = -1.
Complex number
Anything of the form a + bi, with a the real part and b the imaginary part.
Conjugate
a - bi is the conjugate of a + bi. Multiplying them clears the i entirely.
Discriminant
b2 - 4ac. Negative means the two roots are complex.
Complex plane
The plane where the horizontal axis is real and the vertical axis imaginary.

Mistakes to watch for

Forgetting that i2 = -1 flips a sign. In (2+3i)(1-4i) the term -12i2 becomes +12, so the real part is 14, not 2. Leaving the i2 in place, or treating it as +1, loses the whole point of the unit.

Writing √-9 = -3. It is 3i. The minus sign lives under the root and comes out as i; it does not simply move in front.

Multiplying √-4 · √-9 as √36 = 6. Convert first: 2i · 3i = 6i2 = -6. The rule √a√b = √ab fails once the numbers are negative.

Leaving i in a denominator. 52-i is not in standard form. Multiply by the conjugate 2+i to get 2+i, which is.

Reporting only one complex root. Complex roots of a real polynomial always come in conjugate pairs, so finding 3+2i means 3-2i is there too.

Descartes named them "imaginary" as an insult — he thought they were nonsense. The name stuck, which is a pity, because they are no less real than √2.

Your phone uses complex numbers constantly. Every signal it sends is described by one, because a complex number carries size and phase in a single symbol.

Electrical engineers write j instead of i, since i was already taken by current.

Problem 1

Simplify i2.

Worked answer

-1

i2=-1

Problem 2

Simplify i3.

Worked answer

-i

i2 · i = -1 · i = -i

Problem 3

Simplify i12.

Worked answer

1

12 ÷ 4 leaves remainder 0

i0 pattern value is 1

Problem 4

Write √-25 in terms of i.

Worked answer

5i

√-25=√25√-1=5i

Problem 5

Simplify (3+2i)+(5-6i).

Worked answer

8-4i

3+5=8

2i-6i=-4i

Problem 6

Simplify i39.

Worked answer

-i

39 = 4(9)+3

i3=-i

Problem 7

Simplify (7-i)-(2+4i).

Worked answer

5-5i

7-2=5

-i-4i=-5i

Problem 8

Expand (2+i)(3-2i).

Worked answer

8-i

6-4i+3i-2i2

-2i2=+2

8-i

Problem 9

Expand (1+3i)2.

Worked answer

-8+6i

1+6i+9i2

9i2=-9

-8+6i

Problem 10

Give the conjugate of 4-7i.

Worked answer

4+7i

4+7i

Problem 11

Expand (5+2i)(5-2i).

Worked answer

29

25-4i2

-4i2=+4

29

Problem 12

Simplify 103-i.

Worked answer

3+i

Bottom: (3-i)(3+i)=9+1=10

Top: 10(3+i)=30+10i

30+10i10=3+i

Problem 13

Simplify 1+i1-i.

Worked answer

i

Bottom: (1-i)(1+i)=1+1=2

Top: (1+i)2=1+2i-1=2i

2i2=i

Problem 14

How many real roots does x2+2x+5=0 have?

  1. None — both roots are complex
  2. One
  3. Two
  4. Three
Worked answer

None — both roots are complex

4-20=-16

Negative discriminant means no real roots

Problem 15

Solve x2+9=0. Give the positive-imaginary root.

Worked answer

3i

x2=-9

x=± 3i

Problem 16

Simplify √-4 · √-9.

Worked answer

-6

2i · 3i = 6i2

=-6

Multiplying under the roots first would wrongly give +6

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