Imaginary numbers are real
Tonight's idea
Five hundred years ago, negative numbers were called "absurd". They stopped being absurd when they became useful. The same thing happened again.
Nothing squared gives a negative, because positive times positive and negative times negative are both positive. So define a new number \(i\) with the single property \(i^2 = -1\), and see what happens if you follow the ordinary rules of algebra.
Numbers like \(3 + 2i\) are called complex numbers: an ordinary part and an \(i\) part. Adding is easy, just add the parts: \((3 + 2i) + (1 + 5i) = 4 + 7i\). To multiply, expand the brackets the usual way, every term in the first times every term in the second, then replace \(i^2\) by \(-1\).
The picture is the real surprise. Plot \(a + bi\) as the point \((a, b)\): ordinary numbers along the horizontal axis, \(i\) one step up. Now multiply any point by \(i\) and plot the result. It has rotated a quarter turn anticlockwise. That explains \(i^2 = -1\) perfectly: two quarter turns make a half turn, and a half turn takes 1 to \(-1\). There was never anything imaginary about it. It was a rotation all along. Try it at mathdetours.com/demos/complex.
Watch one: multiply (2 + 3i)(1 − i)
Expand like any pair of brackets: \(2 \times 1 + 2 \times (-i) + 3i \times 1 + 3i \times (-i) = 2 - 2i + 3i - 3i^2\).
Now use \(i^2 = -1\): \(-3i^2 = +3\). So the total is \(2 + 3 + (-2i + 3i) = 5 + i\).
Your turn
- Work out \(i^1, i^2, i^3, i^4, i^5\). You should see a cycle. Use it to find \(i^7\) and \(i^{100}\).
- Compute \((2 + 3i) + (1 - i)\) and \((2 + 3i) - (1 - i)\).
- Compute \((1 + 2i)(3 + i)\). Show the \(i^2\) step.
- Compute \((3 + 2i)(3 - 2i)\). The \(i\) parts should vanish. Then compute \((a + bi)(a - bi)\) in general. Notice the answer, 13, equals \(3^2 + 2^2\).
- On a grid, plot \(2 + i\). Multiply it by \(i\) and plot the result. Multiply by \(i\) again, and again, and again, plotting each time. Describe the shape your four points make and say why the fifth point is back where you started.
- Solve \(x^2 = -9\) and \(x^2 = -2\). (Each has two answers.)
Stretch: an equation with no ordinary solution
The equation \(x^2 - 2x + 5 = 0\) has no ordinary solution. Try a few values of \(x\) and notice the left side never reaches zero. Now rewrite it: \(x^2 - 2x + 5 = (x - 1)^2 + 4\). (Check that by expanding.) So the equation says \((x - 1)^2 = -4\). Finish it off with \(i\). Then check one of your answers by substituting it back in.
And a picture question: \(1 + i\) sits at 45° from the real axis, at distance \(\sqrt{2}\) from the centre. Compute \((1 + i)^2\). Where does it land, at what angle and what distance? Can you guess the rule for what squaring does to the angle and to the distance?
Answers
- \(i, -1, -i, 1, i\): a cycle of length 4. \(i^7 = i^3 = -i\) (since \(7 = 4 + 3\)). \(i^{100} = 1\) (100 is a multiple of 4).
- \(3 + 2i\) and \(1 + 4i\).
- \(3 + i + 6i + 2i^2 = 3 + 7i - 2 = 1 + 7i\).
- \(9 - 6i + 6i - 4i^2 = 9 + 4 = 13\). In general \((a + bi)(a - bi) = a^2 + b^2\). So \(13 = 3^2 + 2^2\).
- \(2 + i \to -1 + 2i \to -2 - i \to 1 - 2i \to 2 + i\). The four points are the corners of a square centred at 0, each a quarter turn from the last. Four quarter turns make a full turn, which is also the statement \(i^4 = 1\).
- \(x = 3i\) or \(x = -3i\) (check: \((3i)^2 = 9i^2 = -9\)). \(x = i\sqrt{2}\) or \(x = -i\sqrt{2}\).
Stretch. \((x - 1)^2 = -4\) means \(x - 1 = 2i\) or \(x - 1 = -2i\), so \(x = 1 + 2i\) or \(x = 1 - 2i\). Check with \(x = 1 + 2i\): \(x^2 = 1 + 4i + 4i^2 = -3 + 4i\), then \(x^2 - 2x + 5 = (-3 + 4i) - (2 + 4i) + 5 = 0\). Picture: \((1 + i)^2 = 1 + 2i + i^2 = 2i\), which sits straight up the \(i\) axis at 90°, at distance 2. The angle doubled (45° to 90°) and the distance squared (\(\sqrt{2}\) to 2). That is the general rule: multiplying complex numbers adds their angles and multiplies their distances. It is the reason they are so good at describing anything that rotates or oscillates.