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The complex plane

Numbers · level 7, lesson 4 of 6

A question to start

Take the complex number \(3 + i\) and multiply it by \(i\). Multiply the answer by \(i\) again, and again:

\[3 + i \;\xrightarrow{\times i}\; -1 + 3i \;\xrightarrow{\times i}\; -3 - i \;\xrightarrow{\times i}\; 1 - 3i \;\xrightarrow{\times i}\; 3 + i\]

Now plot these four as points, using the real part across and the imaginary part up: \((3, 1)\), \((-1, 3)\), \((-3, -1)\), \((1, -3)\). They are the four corners of a perfect square centred at 0. Multiplying by \(i\), a piece of pure algebra, has turned the point round by a quarter turn each time. This lesson explains why, and shows that every operation on complex numbers is also a movement in a plane.

The idea

A plane of numbers

The real numbers fill a line. A complex number \(a + bi\) needs two real numbers, so it needs a plane: go \(a\) across and \(b\) up. This picture is called the complex plane (or an Argand diagram, after Jean-Robert Argand, who published it in 1806; Caspar Wessel had described the same idea in 1799).

  • The horizontal axis is the real axis. The real numbers live on it.
  • The vertical axis is the imaginary axis. The imaginary numbers \(bi\) live on it.
  • The number 0 is the origin, where the axes cross.

It is often useful to think of \(z\) not just as a point, but as an arrow from 0 to that point. An arrow with a length and a direction is called a vector.

Adding is moving

Adding complex numbers adds the real parts and the imaginary parts separately, which is exactly how you put two arrows nose to tail. To find \(z + w\), start at 0, follow the arrow \(z\), then from where it ends follow the arrow \(w\).

z = 3 + i w = 1 + 2i z + w = 4 + 3i real imaginary

Going \(z\) then \(w\), or \(w\) then \(z\), ends at the same place: the two routes are the two sides of a parallelogram, and \(z + w\) is its diagonal. This is the parallelogram rule for adding vectors, and it is why complex numbers are used to add forces and velocities.

Subtracting. \(w - z\) is the number you add to \(z\) to reach \(w\). As an arrow, it goes from \(z\) to \(w\). So the distance between two complex numbers \(z\) and \(w\) is the length of \(w - z\).

The modulus: distance from 0

The modulus of \(z = a + bi\), written \(|z|\), is the length of its arrow, the distance from 0 to the point \((a, b)\). The arrow is the long side of a right triangle with legs \(a\) and \(b\), so by Pythagoras

\[|a + bi| = \sqrt{a^2 + b^2}\]

For a real number, \(|a + 0i| = \sqrt{a^2}\), which is the usual absolute value: \(|-5| = 5\). So the modulus is the absolute value, extended to the plane. Comparing with the last lesson, \(z\,\bar z = a^2 + b^2\), so

\[|z|^2 = z\,\bar z\]

The distance between \(z\) and \(w\) is \(|w - z|\). For \(z = 1 + 2i\) and \(w = 4 - 2i\): \(w - z = 3 - 4i\), and the distance is \(\sqrt{9 + 16} = 5\). This is the distance formula from coordinate geometry, in new clothes.

The modulus of a product is the product of the moduli. For any \(z\) and \(w\):

\[|zw| = |z|\,|w|\]

Proof, using \(|z|^2 = z\bar z\) and \(\overline{zw} = \bar z\,\bar w\) from the last lesson:

\[|zw|^2 = (zw)\,\overline{(zw)} = z\,w\,\bar z\,\bar w = (z\,\bar z)(w\,\bar w) = |z|^2\,|w|^2\]

Both sides are lengths (not negative), so taking square roots gives \(|zw| = |z|\,|w|\). ∎

Written out with \(z = a + bi\) and \(w = c + di\), this says something surprising about ordinary whole numbers:

\[(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2\]

A sum of two squares times a sum of two squares is again a sum of two squares. For example \(5 = 1^2 + 2^2\) and \(13 = 2^2 + 3^2\), and \((1 + 2i)(2 + 3i) = -4 + 7i\), so \(65 = 4^2 + 7^2\). Brahmagupta knew this identity in the 600s, long before complex numbers.

Simple movements

Each basic operation moves the point in a simple way.

Operation on \(z = a + bi\) Result Movement
\(z + w\) add parts slide by the arrow \(w\)
\(\bar z\) \(a - bi\) reflect in the real axis
\(-z\) \(-a - bi\) half turn about 0
\(kz\) (\(k > 0\) real) \(ka + kbi\) stretch away from 0 by factor \(k\)
\(iz\) \(-b + ai\) quarter turn anticlockwise about 0

Multiplying by \(i\) is a quarter turn

Why does \(iz\) turn \(z\) by a quarter turn? Work it out: \(i(a + bi) = ai + bi^2 = -b + ai\). So the point \((a, b)\) moves to \((-b, a)\). Two checks:

  1. Same length: \(\sqrt{(-b)^2 + a^2} = \sqrt{a^2 + b^2}\). (Or: \(|iz| = |i|\,|z| = 1 \times |z|\).)
  2. At right angles: the arrow to \((a, b)\) has slope \(\dfrac{b}{a}\) and the arrow to \((-b, a)\) has slope \(\dfrac{a}{-b}\). The product of the slopes is \(-1\), which is the test for perpendicular lines. (If \(a\) or \(b\) is 0, one arrow is on an axis and the other on the other axis, also perpendicular.)

And the turn is anticlockwise: \(1\) (pointing right) goes to \(i\) (pointing up). So multiplying by \(i\) is a rotation of 90° anticlockwise about 0.

3 + i −1 + 3i −3 − i 1 − 3i × i

This is the deep reason why \(i^2 = -1\): two quarter turns make a half turn, and a half turn takes 1 to \(-1\). The mysterious rule from lesson 2 is simply the geometry of turning.

Multiplying by a general complex number does a turn and a stretch at the same time. For example \((2 + i)(1 + i) = 1 + 3i\): the length goes from \(\sqrt5\) to \(\sqrt{10}\) (stretched by \(|1 + i| = \sqrt2\)) and the arrow turns by 45°. Level 8 makes this exact: lengths multiply, angles add.

Circles

The set of all \(z\) with \(|z - c| = r\) is all points at distance \(r\) from the point \(c\): a circle with centre \(c\) and radius \(r\). And \(|z - c| < r\) is the inside of that circle (a disc). This is a compact way to describe regions of the plane, and the real-number version of it, \(|x - c| < r\), is the subject of lesson 6.

Worked examples

1. Modulus. Find \(|3 + 4i|\), \(|-5 + 12i|\), \(|1 - 2i|\) and \(|(3 + 4i)(1 - 2i)|\).

  1. \(|3 + 4i| = \sqrt{9 + 16} = 5\).
  2. \(|-5 + 12i| = \sqrt{25 + 144} = \sqrt{169} = 13\).
  3. \(|1 - 2i| = \sqrt{1 + 4} = \sqrt5\).
  4. Without multiplying: \(|(3 + 4i)(1 - 2i)| = 5\sqrt5\). Check by multiplying: \((3 + 4i)(1 - 2i) = 3 - 6i + 4i - 8i^2 = 11 - 2i\), and \(\sqrt{121 + 4} = \sqrt{125} = 5\sqrt5\). ✓

2. Distance. How far apart are \(-2 + i\) and \(4 - 7i\)?

\(w - z = (4 - (-2)) + (-7 - 1)i = 6 - 8i\), so the distance is \(\sqrt{36 + 64} = 10\).

3. The fourth corner. A parallelogram has corners \(0\), \(2 + i\) and \(1 + 3i\), with 0 opposite the missing corner. Find it.

The missing corner is the end of the diagonal from 0, so it is the sum: \((2 + i) + (1 + 3i) = 3 + 4i\).

4. A square from one corner. A square is centred at 0 and one corner is \(2 + 5i\). Find the other three corners.

Each corner is a quarter turn of the last, so multiply by \(i\) repeatedly:

  1. \(i(2 + 5i) = 2i + 5i^2 = -5 + 2i\).
  2. \(i(-5 + 2i) = -5i + 2i^2 = -2 - 5i\). (This is \(-(2 + 5i)\), the half turn.)
  3. \(i(-2 - 5i) = -2i - 5i^2 = 5 - 2i\).

5. Describe a set. Describe all \(z\) with \(|z - 1 - i| = 2\), and all \(z\) with \(|z - 3| = |z + 1|\).

  1. \(|z - (1 + i)| = 2\): the points at distance 2 from \(1 + i\), a circle with centre \((1, 1)\) and radius 2.
  2. \(|z - 3| = |z - (-1)|\): the points the same distance from \(3\) and from \(-1\). That is the perpendicular bisector of the segment from \(-1\) to \(3\): the vertical line through \(1\), all numbers \(1 + bi\).

Try it

  1. Plot \(2 + 3i\), its conjugate, its negative, and \(i\) times it. What shape do the four points make?
  2. Find \(|6 - 8i|\), \(|-7i|\) and \(|\tfrac35 + \tfrac45 i|\).
  3. Find the distance between \(1 + 5i\) and \(-4 - 7i\).
  4. Without multiplying out, find \(|(1 + i)^{10}|\).
  5. Multiply \(4 - i\) by \(-i\). What movement does multiplying by \(-i\) do? Explain using \(-i = i^3\).
  6. Write \(10 = 1^2 + 3^2\) and \(13 = 2^2 + 3^2\). Use complex numbers to write \(130\) as a sum of two squares.
  7. Describe the set of all \(z\) with \(|z + 2i| \le 3\).
  8. Show that \(|\bar z| = |z|\) and \(|z + w|^2 = |z|^2 + |w|^2 + z\bar w + \bar z w\). (Hint for the second: write \(|z + w|^2 = (z + w)\overline{(z + w)}\).)
Answers
  1. \(2 + 3i\) at \((2, 3)\); conjugate \(2 - 3i\) at \((2, -3)\); negative \(-2 - 3i\) at \((-2, -3)\); \(i(2 + 3i) = -3 + 2i\) at \((-3, 2)\). The first three are three corners of a rectangle (with \(-2 + 3i\) it would be complete). The point \(-3 + 2i\) is \(2 + 3i\) turned a quarter turn. All four are \(\sqrt{13}\) from 0, so they lie on one circle.
  2. \(\sqrt{36 + 64} = 10\). \(|-7i| = 7\). \(\sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = 1\).
  3. \((-4 - 7i) - (1 + 5i) = -5 - 12i\), distance \(\sqrt{25 + 144} = 13\).
  4. \(|1 + i| = \sqrt2\), and moduli multiply, so \(|(1 + i)^{10}| = (\sqrt2)^{10} = 2^5 = 32\).
  5. \(-i(4 - i) = -4i + i^2 = -1 - 4i\). Multiplying by \(-i = i^3\) is three quarter turns anticlockwise, which is the same as one quarter turn clockwise. Check: \((4, -1)\) to \((-1, -4)\) is a clockwise quarter turn.
  6. \((1 + 3i)(2 + 3i) = 2 + 3i + 6i + 9i^2 = -7 + 9i\), so \(130 = 7^2 + 9^2 = 49 + 81\). ✓ (Using the conjugate \((1 + 3i)(2 - 3i) = 2 - 3i + 6i + 9 = 11 + 3i\) gives another way: \(130 = 11^2 + 3^2 = 121 + 9\). ✓)
  7. \(|z - (-2i)| \le 3\): all points at distance at most 3 from \(-2i\), the point \((0, -2)\). A filled disc with centre \((0, -2)\) and radius 3, including its edge.
  8. \(\bar z = a - bi\) has modulus \(\sqrt{a^2 + (-b)^2} = |z|\). For the second, \((z + w)(\bar z + \bar w) = z\bar z + z\bar w + w\bar z + w\bar w = |z|^2 + |w|^2 + z\bar w + \bar z w\). (The last two terms are conjugates of each other, so their sum is real: \(2\operatorname{Re}(z\bar w)\).)

Common mistake

Taking the modulus of each part separately. \(|3 - 4i|\) is not \(3 + 4\) or \(3 - 4\). It is a length, found by Pythagoras: \(\sqrt{3^2 + 4^2} = 5\). And it is never negative.

Thinking \(|z + w| = |z| + |w|\). Lengths of arrows do not simply add, any more than two sides of a triangle add to the third. \(|3| + |4i| = 7\), but \(|3 + 4i| = 5\). They are equal only when the arrows point the same way. In lesson 6 this becomes the triangle inequality, \(|z + w| \le |z| + |w|\).

Where it leads