Skip to content

The number i

Numbers · level 7, lesson 2 of 6

A question to start

Here is an equation with an obvious answer:

\[x^3 = 15x + 4\]

Try \(x = 4\): the left side is \(64\) and the right side is \(60 + 4 = 64\). It works.

In the 1500s, Italian mathematicians found a formula for equations like this one, a cubic formula, the way the quadratic formula solves \(ax^2 + bx + c = 0\). Put this equation into it and the formula says

\[x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}\]

The square root of a negative number, in the middle of the working, for an equation whose answer is the ordinary number 4. Either the formula is broken, or the square root of a negative number means something. By the end of this lesson you will be able to work this out and get exactly 4.

The idea

Defining a new number

In lesson 1 we saw that no real number squares to a negative. So we do what was done for negatives and fractions: we invent a number with the property we want, and insist that the usual rules of arithmetic still hold.

The number \(i\) is a number with \(i^2 = -1\).

That is the whole definition. Everything else comes from the ordinary rules: adding, multiplying, brackets, the distributive law. The letter \(i\) stands for imaginary. René Descartes called such roots "imaginary" in 1637, and the name stuck long after anyone thought of them as fake. The letter \(i\) itself was first used by Leonhard Euler, late in the 1700s.

Square roots of negative numbers

Once we have \(i\), every negative number has square roots.

\((3i)^2 = 3^2 \times i^2 = 9 \times (-1) = -9\). And \((-3i)^2 = (-3)^2 \times i^2 = -9\) as well. So \(-9\) has two square roots, \(3i\) and \(-3i\), just as \(9\) has two square roots, \(3\) and \(-3\).

As with positive numbers, the symbol \(\sqrt{\phantom{x}}\) picks one of them. For a positive number \(a\) we agree that

\[\sqrt{-a} = i\sqrt{a}\]

so \(\sqrt{-9} = 3i\), \(\sqrt{-2} = i\sqrt2\), \(\sqrt{-1} = i\). Notice that \(-i\) also squares to \(-1\): \((-i)^2 = (-1)^2 i^2 = -1\). The two numbers \(i\) and \(-i\) are completely alike in their arithmetic; calling one of them "\(i\)" is a choice, like choosing which way along the number line is positive.

Numbers of the form \(bi\), with \(b\) real, are called imaginary numbers (or "pure imaginary"): \(2i\), \(-7i\), \(i\sqrt3\), \(\tfrac12 i\). In lesson 3 we add real numbers to them and get the complex numbers.

The powers of \(i\) go round in a cycle

Multiply by \(i\) again and again, using only \(i^2 = -1\):

power \(i^1\) \(i^2\) \(i^3\) \(i^4\) \(i^5\) \(i^6\) \(i^7\) \(i^8\)
working \(i\) \(-1\) \(i^2 \cdot i\) \(i^2 \cdot i^2\) \(i^4 \cdot i\) \(i^4 \cdot i^2\) \(i^4 \cdot i^3\) \(i^4 \cdot i^4\)
value \(i\) \(-1\) \(-i\) \(1\) \(i\) \(-1\) \(-i\) \(1\)

Once \(i^4 = 1\), the pattern must repeat, because \(i^{n+4} = i^n \times i^4 = i^n\). So the powers of \(i\) run through four values forever: \(i, -1, -i, 1, i, -1, -i, 1, \ldots\)

1 i −1 −i × i × i × i × i

The picture is a preview of lesson 4: if \(1\) sits to the right and \(i\) sits straight up, each multiplication by \(i\) is a quarter turn, and four quarter turns bring you home.

The rule for any power. Divide the exponent by 4 and look only at the remainder:

remainder of \(n \div 4\) 0 1 2 3
\(i^n\) \(1\) \(i\) \(-1\) \(-i\)

For example \(i^{27}\): \(27 = 4 \times 6 + 3\), so \(i^{27} = (i^4)^6 \times i^3 = 1 \times (-i) = -i\).

Negative powers too. What is \(\dfrac{1}{i}\)? Ask what times \(i\) gives 1. Since \(i \times (-i) = -i^2 = 1\), the answer is \(\dfrac1i = -i\). So \(i^{-1} = -i\), which is also \(i^3\), exactly what the cycle predicts if you keep going backwards: \(\ldots, i^{-1} = -i, \; i^0 = 1, \; i^1 = i, \ldots\) The rule "remainder after dividing by 4" works for negative \(n\) as long as you use a remainder of 0, 1, 2 or 3: \(-1 = 4 \times (-1) + 3\), remainder 3, giving \(-i\).

Four in a row add to zero. \(i + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0\). Any four consecutive powers are this same block multiplied by some power of \(i\), so they also add to 0.

The honest history: cubic equations, not quadratics

You might guess that \(\sqrt{-1}\) was invented to solve \(x^2 + 1 = 0\). It was not. For a quadratic with no real solutions, people simply said "no solution", and that is a perfectly reasonable answer when a quadratic describes a real situation.

Cardano's puzzle (1545). In his book Ars Magna ("The Great Art"), Gerolamo Cardano asked for two numbers that add to 10 and multiply to 40. The equations lead to \(5 + \sqrt{-15}\) and \(5 - \sqrt{-15}\). Cardano checked that they work, calling the calculation as subtle as it is useless:

\[\big(5 + \sqrt{-15}\big) + \big(5 - \sqrt{-15}\big) = 10, \qquad \big(5 + \sqrt{-15}\big)\big(5 - \sqrt{-15}\big) = 25 - (-15) = 40.\]

The cubic formula forced the issue. The same book published a formula (found by Scipione del Ferro and Niccolò Tartaglia) for cubics of the form \(x^3 = px + q\):

\[x = \sqrt[3]{\frac q2 + \sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} + \sqrt[3]{\frac q2 - \sqrt{\frac{q^2}{4} - \frac{p^3}{27}}}\]

For \(x^3 = 15x + 4\) we have \(p = 15\), \(q = 4\), so \(\dfrac{q^2}{4} = 4\) and \(\dfrac{p^3}{27} = 125\), and the square root is \(\sqrt{4 - 125} = \sqrt{-121}\). This is the equation from the start of the lesson. Here "no solution" is plainly wrong: \(x = 4\) works. The square roots of negatives could not be thrown away.

Bombelli's answer (1572). In his Algebra, Rafael Bombelli had the bold idea of simply calculating with these roots by the usual rules. He noticed that

\[(2 + i)^3 = 2 + 11i \qquad\text{and}\qquad (2 - i)^3 = 2 - 11i\]

(you will check the first in worked example 4). Since \(\sqrt{-121} = 11i\), the formula becomes

\[x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i} = (2 + i) + (2 - i) = 4.\]

The imaginary parts cancel and the true answer appears. This is the moment complex numbers became necessary: a route through them led to a real answer that no route around them could reach with that formula. It took another two hundred years, and the work of Euler and Gauss, before they were fully accepted.

Worked examples

1. Powers of \(i\). Find \(i^{50}\), \(i^{100}\), \(i^{2027}\) and \(i^{-3}\).

  1. \(50 = 4 \times 12 + 2\), remainder 2: \(i^{50} = -1\).
  2. \(100 = 4 \times 25\), remainder 0: \(i^{100} = 1\).
  3. Only the last two digits matter for the remainder after dividing by 4 (because 100 is a multiple of 4). \(27 = 4 \times 6 + 3\), remainder 3: \(i^{2027} = -i\).
  4. \(i^{-3} = \dfrac{1}{i^3} = \dfrac{1}{-i}\). Since \(\dfrac1i = -i\), we get \(\dfrac{1}{-i} = i\). Check with the cycle: \(-3 = 4 \times (-1) + 1\), remainder 1, giving \(i\). ✓

2. Square roots of negatives. Simplify \(\sqrt{-50}\) and \(\sqrt{-\tfrac{4}{9}}\).

  1. \(\sqrt{-50} = i\sqrt{50} = i\sqrt{25 \times 2} = 5i\sqrt2\). Check: \((5i\sqrt2)^2 = 25 \times i^2 \times 2 = -50\). ✓
  2. \(\sqrt{-\tfrac49} = i\sqrt{\tfrac49} = \tfrac23 i\). Check: \(\left(\tfrac23 i\right)^2 = \tfrac49 \times (-1)\). ✓

3. Multiplying roots of negatives. Work out \(\sqrt{-4} \times \sqrt{-9}\).

Write each with \(i\) first: \(\sqrt{-4} = 2i\) and \(\sqrt{-9} = 3i\). Then \(2i \times 3i = 6i^2 = -6\).

The tempting shortcut \(\sqrt{-4}\times\sqrt{-9} = \sqrt{(-4)(-9)} = \sqrt{36} = 6\) gives the wrong sign. See the common mistake below.

4. Bombelli's check. Show that \((2 + i)^3 = 2 + 11i\).

  1. First square: \((2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i\).
  2. Then multiply by \(2 + i\) once more: \((3 + 4i)(2 + i) = 6 + 3i + 8i + 4i^2 = 6 + 11i - 4 = 2 + 11i\). ✓

Changing every \(i\) to \(-i\) in the same working gives \((2 - i)^3 = 2 - 11i\).

5. Solving with \(i\). Solve \(x^2 + 16 = 0\) and \(3x^2 + 75 = 0\).

  1. \(x^2 = -16\), so \(x = \pm\sqrt{-16} = \pm 4i\). Check: \((4i)^2 + 16 = -16 + 16 = 0\). ✓
  2. \(3x^2 = -75\), \(x^2 = -25\), so \(x = \pm 5i\).

Try it

  1. Find \(i^6\), \(i^{11}\), \(i^{40}\) and \(i^{93}\).
  2. Simplify \(\sqrt{-36}\), \(\sqrt{-7}\) and \(\sqrt{-12}\).
  3. Work out \((3i)^2\), \((-2i)^3\) and \(i^3 \times i^5\).
  4. Work out \(\sqrt{-2}\times\sqrt{-8}\) correctly.
  5. Solve \(x^2 = -81\) and \(2x^2 + 18 = 0\).
  6. Find \(\dfrac{1}{i^2}\), \(\dfrac{1}{i^3}\) and \(i^{-10}\).
  7. Find the sum \(i + i^2 + i^3 + \cdots + i^{102}\).
  8. Check Cardano's numbers another way: show that \(x = 5 + \sqrt{-15}\) solves \(x^2 - 10x + 40 = 0\).
Answers
  1. \(6 = 4 + 2\): \(-1\). \(11 = 8 + 3\): \(-i\). \(40 = 4 \times 10\): \(1\). \(93 = 92 + 1\): \(i\).
  2. \(\sqrt{-36} = 6i\). \(\sqrt{-7} = i\sqrt7\). \(\sqrt{-12} = i\sqrt{12} = 2i\sqrt3\).
  3. \((3i)^2 = 9i^2 = -9\). \((-2i)^3 = (-2)^3 i^3 = -8 \times (-i) = 8i\). \(i^3 \times i^5 = i^8 = 1\).
  4. \(\sqrt{-2} = i\sqrt2\) and \(\sqrt{-8} = i\sqrt8 = 2i\sqrt2\). Product: \(i\sqrt2 \times 2i\sqrt2 = 2 \times 2 \times i^2 = -4\). (Not \(\sqrt{16} = 4\).)
  5. \(x = \pm 9i\). And \(2x^2 = -18\), \(x^2 = -9\), \(x = \pm 3i\).
  6. \(\dfrac{1}{i^2} = \dfrac{1}{-1} = -1\). \(\dfrac{1}{i^3} = \dfrac{1}{-i} = i\) (since \(i \times (-i) = 1\), so \(\dfrac{1}{-i} = i\)). \(i^{-10}\): \(-10 = 4 \times (-3) + 2\), remainder 2, so \(-1\). Or \(i^{-10} = \dfrac{1}{i^{10}} = \dfrac{1}{-1} = -1\).
  7. The first 100 powers make 25 blocks of four, each adding to 0. What is left is \(i^{101} + i^{102} = i + (-1) = -1 + i\).
  8. \(x^2 = 25 + 10\sqrt{-15} + (\sqrt{-15})^2 = 25 + 10\sqrt{-15} - 15 = 10 + 10\sqrt{-15}\). Then \(x^2 - 10x + 40 = 10 + 10\sqrt{-15} - 50 - 10\sqrt{-15} + 40 = 0\). ✓

Common mistake

Using \(\sqrt a \times \sqrt b = \sqrt{ab}\) when \(a\) and \(b\) are both negative. It would "prove"

\[-1 = i \times i = \sqrt{-1}\times\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1,\]

which is false. The rule \(\sqrt a\sqrt b = \sqrt{ab}\) was proved for \(a, b \ge 0\) and does not carry over. The safe habit: always rewrite \(\sqrt{-a}\) as \(i\sqrt a\) first, then multiply.

Getting the cycle wrong by starting at \(i^0\). The remainder 1 gives \(i\) and remainder 0 gives 1. Students sometimes count "first value, second value, ..." and slip by one. A quick check: \(i^4 = 1\), and 4 has remainder 0.

Where it leads