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Polar form

Numbers · level 8, lesson 1 of 6

A question to start

Multiply these two complex numbers the ordinary way:

\[(1 + i\sqrt3)(\sqrt3 + i) = \sqrt3 + i + 3i + i^2\sqrt3 = \sqrt3 + 4i - \sqrt3 = 4i.\]

The real parts cancel exactly and the answer is the pure imaginary number \(4i\). Why so tidy?

Draw them. The arrow to \(1 + i\sqrt3\) has length 2 and points 60° above the real axis. The arrow to \(\sqrt3 + i\) has length 2 and points 30° up. The answer \(4i\) has length \(2 \times 2 = 4\) and points \(60° + 30° = 90°\) up. The lengths multiplied and the angles added. This lesson shows that this always happens, and proves it.

The idea

Length and angle

In level 7 you located \(z = a + bi\) by going \(a\) across and \(b\) up: its rectangular (or Cartesian) form. You can also locate the same point by saying how far it is from 0 and in which direction:

  • the modulus \(r = |z| = \sqrt{a^2 + b^2}\), its distance from 0;
  • the argument \(\theta = \arg z\), the angle from the positive real axis to the arrow, measured anticlockwise.

This is like giving directions as "5 km on a bearing of 30°" instead of "4.3 km east, 2.5 km north". Both describe the same place.

Radians. From here on, angles are usually measured in radians: an angle of 1 radian at the centre of a circle cuts off an arc as long as the radius. A full turn is \(2\pi\) radians, so

\[\pi \text{ radians} = 180°, \qquad \tfrac{\pi}{2} = 90°, \qquad \tfrac{\pi}{3} = 60°, \qquad \tfrac{\pi}{4} = 45°, \qquad \tfrac{\pi}{6} = 30°.\]

Radians are not just another unit; they are the natural one, and Euler's formula in lesson 4 only works with them. You may use degrees for drawing, but switch to radians for calculating.

The argument is not unique. Turning a full extra turn lands on the same arrow, so \(\theta\), \(\theta + 2\pi\), \(\theta - 2\pi\), ... are all arguments of the same \(z\). To have one definite answer we use the principal argument, the one with \(-\pi < \theta \le \pi\) (between \(-180°\) and \(180°\), including \(180°\)). The number 0 has no argument at all: an arrow of length 0 points nowhere.

From polar to rectangular

Draw the right triangle under the arrow: hypotenuse \(r\), angle \(\theta\) at 0. By the definitions of cosine and sine,

\[a = r\cos\theta, \qquad b = r\sin\theta.\]

(These hold in every quadrant, because that is how cosine and sine are defined for all angles: the coordinates of the point at angle \(\theta\) on the circle of radius 1.) So

\[z = a + bi = r(\cos\theta + i\sin\theta)\]

This is the polar form (or modulus-argument form) of \(z\). The bracket is written so often that it has a short name: \(\operatorname{cis}\theta = \cos\theta + i\sin\theta\) ("c-i-s"). Then \(z = r\operatorname{cis}\theta\). Note that \(\operatorname{cis}\theta\) itself always has modulus \(\sqrt{\cos^2\theta + \sin^2\theta} = 1\): it is a point on the unit circle, and \(r\) stretches it out to length \(r\).

From rectangular to polar

Given \(a + bi\):

  1. \(r = \sqrt{a^2 + b^2}\).
  2. For the angle, \(\tan\theta = \dfrac ba\). But a calculator's \(\tan^{-1}\) only returns angles between \(-90°\) and \(90°\), which is the right half of the plane. Always plot the point first and choose the angle in the correct quadrant.

A reliable method: find the reference angle \(\alpha = \tan^{-1}\left|\dfrac ba\right|\) (an acute angle), then

Quadrant Signs Principal argument
first \(a > 0\), \(b > 0\) \(\theta = \alpha\)
second \(a < 0\), \(b > 0\) \(\theta = \pi - \alpha\)
third \(a < 0\), \(b < 0\) \(\theta = -(\pi - \alpha)\)
fourth \(a > 0\), \(b < 0\) \(\theta = -\alpha\)

Points on the axes are easy to read off: \(\arg 5 = 0\), \(\arg(3i) = \tfrac\pi2\), \(\arg(-2) = \pi\), \(\arg(-i) = -\tfrac\pi2\).

Multiplying: lengths multiply, angles add

Theorem. If \(z = r\operatorname{cis}\alpha\) and \(w = s\operatorname{cis}\beta\), then

\[zw = rs\operatorname{cis}(\alpha + \beta).\]

Proof. The real numbers \(r\) and \(s\) just multiply, so the work is in the brackets. Expand as in level 7:

\[(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta).\]

Now recall the angle-sum formulas from trigonometry:

\[\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta, \qquad \sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.\]

The real part is exactly \(\cos(\alpha + \beta)\) and the imaginary part is exactly \(\sin(\alpha + \beta)\). So \(\operatorname{cis}\alpha\,\operatorname{cis}\beta = \operatorname{cis}(\alpha + \beta)\), and multiplying by \(rs\) finishes the proof. ∎

Read the proof the other way round and it is just as interesting: complex multiplication is the angle-sum formulas in disguise. If you ever forget them, multiply \(\operatorname{cis}\alpha\) by \(\operatorname{cis}\beta\) and read them off.

√3 + i: length 2, 30° 1 + i√3: length 2, 60° 4i: length 4, 90° real

Dividing, reciprocals and conjugates

Division undoes multiplication, so it divides the lengths and subtracts the angles:

\[\frac{r\operatorname{cis}\alpha}{s\operatorname{cis}\beta} = \frac rs\operatorname{cis}(\alpha - \beta) \qquad (s \ne 0).\]

(Check: multiply the right side by \(s\operatorname{cis}\beta\); lengths give \(r\), angles give \(\alpha\).) In particular

\[\frac{1}{r\operatorname{cis}\theta} = \frac1r\operatorname{cis}(-\theta), \qquad \overline{r\operatorname{cis}\theta} = r\operatorname{cis}(-\theta).\]

The conjugate keeps the length and flips the angle, which is the reflection in the real axis you met in level 7.

What multiplication does to the whole plane

Fix \(w = s\operatorname{cis}\beta\). The rule "multiply every complex number by \(w\)" moves every point of the plane: each arrow is turned by \(\beta\) and stretched by \(s\). That is one rotation and one scaling, both about 0, so shapes keep their shape: a square goes to a square, a triangle to a similar triangle. Multiplying by \(i = \operatorname{cis}\tfrac\pi2\) is the quarter turn of level 7, with \(s = 1\). This is why engineers and physicists use complex numbers for anything that rotates or oscillates: rotating is just multiplying.

Worked examples

1. Rectangular to polar (second quadrant). Write \(-1 + i\sqrt3\) in polar form.

  1. \(r = \sqrt{1 + 3} = 2\).
  2. The point \((-1, \sqrt3)\) is in the second quadrant. Reference angle: \(\tan\alpha = \dfrac{\sqrt3}{1}\), so \(\alpha = \tfrac\pi3\).
  3. \(\theta = \pi - \tfrac\pi3 = \tfrac{2\pi}{3}\) (that is, 120°).
  4. So \(-1 + i\sqrt3 = 2\operatorname{cis}\tfrac{2\pi}3\). Check: \(2\cos\tfrac{2\pi}{3} = 2 \times (-\tfrac12) = -1\) ✓ and \(2\sin\tfrac{2\pi}{3} = 2 \times \tfrac{\sqrt3}{2} = \sqrt3\) ✓.

2. Rectangular to polar (third quadrant). Write \(-3 - 4i\) in polar form.

  1. \(r = \sqrt{9 + 16} = 5\).
  2. Third quadrant. Reference angle \(\alpha = \tan^{-1}\tfrac43 \approx 0.9273\) rad (about 53.13°).
  3. \(\theta = -(\pi - 0.9273) \approx -2.2143\) rad (about \(-126.87°\)).
  4. A calculator's \(\tan^{-1}\left(\tfrac{-4}{-3}\right) = \tan^{-1}\tfrac43 \approx 0.9273\) would put the point in the first quadrant, at \(3 + 4i\): wrong by a half turn.

3. Polar to rectangular. Write \(4\operatorname{cis}\left(-\tfrac\pi6\right)\) as \(a + bi\).

\(a = 4\cos\left(-\tfrac\pi6\right) = 4 \times \tfrac{\sqrt3}{2} = 2\sqrt3\) and \(b = 4\sin\left(-\tfrac\pi6\right) = 4 \times \left(-\tfrac12\right) = -2\). So the number is \(2\sqrt3 - 2i\).

4. An exact value of \(\cos 75°\). Multiply \(\operatorname{cis}45°\) by \(\operatorname{cis}30°\) in both forms.

  1. Polar: the product is \(\operatorname{cis}75°\).
  2. Rectangular: \(\left(\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}i\right)\left(\tfrac{\sqrt3}{2} + \tfrac12 i\right) = \tfrac{\sqrt6}{4} + \tfrac{\sqrt2}{4}i + \tfrac{\sqrt6}{4}i + \tfrac{\sqrt2}{4}i^2 = \tfrac{\sqrt6 - \sqrt2}{4} + \tfrac{\sqrt6 + \sqrt2}{4}i\).
  3. Equal complex numbers have equal parts, so \(\cos75° = \tfrac{\sqrt6 - \sqrt2}{4} \approx 0.2588\) and \(\sin75° = \tfrac{\sqrt6 + \sqrt2}{4} \approx 0.9659\). (A calculator agrees.)

5. Dividing in polar form. Find \(\dfrac{-1 + i\sqrt3}{1 + i}\) in polar form, then in rectangular form.

  1. Top: \(2\operatorname{cis}120°\) (example 1). Bottom: \(1 + i = \sqrt2\operatorname{cis}45°\).
  2. Quotient: \(\dfrac{2}{\sqrt2}\operatorname{cis}(120° - 45°) = \sqrt2\operatorname{cis}75°\).
  3. Using example 4: \(\sqrt2 \times \tfrac{\sqrt6 - \sqrt2}{4} = \tfrac{\sqrt{12} - 2}{4} = \tfrac{\sqrt3 - 1}{2}\) for the real part, and \(\sqrt2 \times \tfrac{\sqrt6 + \sqrt2}{4} = \tfrac{\sqrt3 + 1}{2}\) for the imaginary part.
  4. Check the level 7 way: multiply top and bottom by \(1 - i\). Top: \((-1 + i\sqrt3)(1 - i) = -1 + i + i\sqrt3 + \sqrt3 = (\sqrt3 - 1) + (\sqrt3 + 1)i\). Bottom: 2. Same answer. ✓

Try it

  1. Write in polar form, with the principal argument in radians: (a) \(1 - i\); (b) \(-2i\); (c) \(-5\); (d) \(-4 + 4i\).
  2. Write \(\sqrt3 - i\) and \(2 - 2i\sqrt3\) in polar form.
  3. Write \(3\operatorname{cis}\tfrac{5\pi}{6}\) and \(\sqrt2\operatorname{cis}\left(-\tfrac{3\pi}{4}\right)\) as \(a + bi\).
  4. Find \(zw\) and \(\dfrac zw\) in polar form for \(z = 6\operatorname{cis}\tfrac{\pi}{2}\), \(w = 2\operatorname{cis}\tfrac{\pi}{3}\).
  5. Find \(2\operatorname{cis}40° \times 5\operatorname{cis}110°\) and write the answer as \(a + bi\) exactly.
  6. Find the exact value of \(\cos15°\) by working out \(\dfrac{\operatorname{cis}45°}{\operatorname{cis}30°}\) in rectangular form.
  7. The point \(3 + i\) is rotated 90° anticlockwise about 0 and then its distance from 0 is doubled. Which single complex number do you multiply by? Where does the point end up?
  8. Show that \(\arg(z\bar z) = 0\) for every \(z \ne 0\), using polar form, and explain why that agrees with level 7.
Answers
  1. (a) \(r = \sqrt2\), fourth quadrant, \(\theta = -\tfrac\pi4\): \(\sqrt2\operatorname{cis}\left(-\tfrac\pi4\right)\). (b) \(2\operatorname{cis}\left(-\tfrac\pi2\right)\). (c) \(5\operatorname{cis}\pi\). (d) \(r = \sqrt{32} = 4\sqrt2\), second quadrant, \(\theta = \pi - \tfrac\pi4 = \tfrac{3\pi}4\): \(4\sqrt2\operatorname{cis}\tfrac{3\pi}{4}\).
  2. \(\sqrt3 - i\): \(r = 2\), fourth quadrant, \(\tan\alpha = \tfrac{1}{\sqrt3}\) so \(\alpha = \tfrac\pi6\): \(2\operatorname{cis}\left(-\tfrac\pi6\right)\). \(2 - 2i\sqrt3\): \(r = \sqrt{4 + 12} = 4\), \(\tan\alpha = \sqrt3\), \(\alpha = \tfrac\pi3\): \(4\operatorname{cis}\left(-\tfrac\pi3\right)\).
  3. \(3\left(-\tfrac{\sqrt3}{2} + \tfrac12 i\right) = -\tfrac{3\sqrt3}{2} + \tfrac32 i\). And \(\sqrt2\left(-\tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2}i\right) = -1 - i\).
  4. \(zw = 12\operatorname{cis}\tfrac{5\pi}{6}\) (since \(\tfrac\pi2 + \tfrac\pi3 = \tfrac{5\pi}6\)). \(\dfrac zw = 3\operatorname{cis}\tfrac{\pi}{6}\) (since \(\tfrac\pi2 - \tfrac\pi3 = \tfrac\pi6\)).
  5. \(10\operatorname{cis}150° = 10\left(-\tfrac{\sqrt3}{2} + \tfrac12 i\right) = -5\sqrt3 + 5i\).
  6. \(\operatorname{cis}45° \times \overline{\operatorname{cis}30°} = \left(\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}i\right)\left(\tfrac{\sqrt3}{2} - \tfrac12 i\right)\) (dividing by a number of modulus 1 is multiplying by its conjugate). Real part: \(\tfrac{\sqrt6}{4} + \tfrac{\sqrt2}{4}\). So \(\cos15° = \tfrac{\sqrt6 + \sqrt2}{4} \approx 0.9659\). (Imaginary part \(\tfrac{\sqrt6 - \sqrt2}{4} = \sin15°\).)
  7. Turn by \(\tfrac\pi2\) and stretch by 2: multiply by \(2\operatorname{cis}\tfrac\pi2 = 2i\). The point goes to \(2i(3 + i) = 6i + 2i^2 = -2 + 6i\).
  8. If \(z = r\operatorname{cis}\theta\) then \(\bar z = r\operatorname{cis}(-\theta)\), so \(z\bar z = r^2\operatorname{cis}(\theta - \theta) = r^2\operatorname{cis}0 = r^2\). Argument 0 means a positive real number, matching level 7's \(z\bar z = a^2 + b^2 = |z|^2\).

Common mistake

Trusting \(\tan^{-1}\dfrac ba\) for the argument. It cannot tell \(3 + 4i\) from \(-3 - 4i\), because \(\dfrac{4}{3} = \dfrac{-4}{-3}\). Always sketch the point and pick the angle in its quadrant. (Many calculators and programming languages have a two-input function, often called atan2, that does this for you.)

Adding the moduli or multiplying the arguments. In a product the lengths multiply and the angles add, never the other way round. A useful check: \(i \times i\) has length \(1 \times 1 = 1\) and angle \(90° + 90° = 180°\), which is \(-1\). ✓

Where it leads