Chapter 2: The Complex Plane and Polar Form

The Gaussian plane, polar form, and De Moivre's theorem

Introduction (high-school level)

The Complex Plane

A complex number $z = a + bi$ can be regarded as a pair of two real numbers $(a, b)$. This lets us depict complex numbers as points in a plane.

Definition: the complex plane

The complex plane (also called the Gaussian plane) is the plane in which complex numbers are represented as points.

  • Horizontal axis (x-axis): the real axis — represents the real part
  • Vertical axis (y-axis): the imaginary axis — represents the imaginary part

The complex number $z = a + bi$ corresponds to the point $(a, b)$.

Real Imaginary O z = 3 + 2i 3 2i
Figure 1: the complex plane

Basic quantities and their geometric meaning

For $z = a + bi$, the definitions of the basic quantities and their geometric meaning on the complex plane are as follows.

  • Real number: when $b = 0$ (so $z = a$). Geometrically, a point on the real axis.
  • Pure imaginary number: when $a = 0$ and $b \neq 0$ (so $z = bi$). Geometrically, a point on the imaginary axis (excluding the origin).
  • Modulus $|z| = \sqrt{a^2 + b^2}$. Geometrically, the distance from the origin to $z$.
  • Complex conjugate $\bar{z} = a - bi$. Geometrically, the reflection of $z$ across the real axis.

Polar Form

A complex number can be expressed not only in Cartesian coordinates $(a, b)$ but also in polar coordinates $(r, \theta)$.

Definition: polar form

Let $z \neq 0$ be a complex number. The polar form of $z$ is:

$$z = r(\cos\theta + i\sin\theta)$$

where:

  • $r = |z|$: the distance from the origin (the modulus, or radius)
  • $\theta$: the angle from the positive real axis to the point $z$ (the argument)

The argument is written $\mathrm{arg}(z)$; restricting it to $-\pi < \theta \leq \pi$ gives the principal value, sometimes written $\mathrm{Arg}(z)$.

Re Im O z r = |z| θ a = r cos θ b = r sin θ
Figure 2: polar form

Relation between Cartesian and polar coordinates

From $z = a + bi = r(\cos\theta + i\sin\theta)$:

  • $a = r\cos\theta$
  • $b = r\sin\theta$
  • $r = \sqrt{a^2 + b^2}$
  • $\tan\theta = \displaystyle\dfrac{b}{a}$ (when $a \neq 0$)

Example: converting to polar form

Express $z = 1 + i$ in polar form.

  • $r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}$
  • $\tan\theta = \dfrac{1}{1} = 1$, and since $z$ lies in the first quadrant, $\theta = \dfrac{\pi}{4}$

Hence $z = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$.

Example: complex numbers at special angles

$\theta$ Polar form Cartesian form
$0$ $\cos 0 + i\sin 0$ $1$
$\dfrac{\pi}{2}$ $\cos\dfrac{\pi}{2} + i\sin\dfrac{\pi}{2}$ $i$
$\pi$ $\cos\pi + i\sin\pi$ $-1$
$\dfrac{3\pi}{2}$ $\cos\dfrac{3\pi}{2} + i\sin\dfrac{3\pi}{2}$ $-i$

This table lists the angles for $0 \le \theta < 2\pi$. For the principal value ($-\pi < \theta \le \pi$), $\dfrac{3\pi}{2}$ corresponds to $-\dfrac{\pi}{2}$ (the same point $-i$).

Multiplication and Division in Polar Form

Theorem: multiplication in polar form

For $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$:

$$z_1 z_2 = r_1 r_2 \left(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right)$$

That is, the moduli multiply and the arguments add.

Proof

\begin{align*} z_1 z_2 &= r_1 r_2 (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2) \\ &= r_1 r_2 \left[(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)\right] \\ &= r_1 r_2 \left(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right) \end{align*}

The last step uses the angle-addition formulas for sine and cosine.

Theorem: division in polar form

For $z_2 \neq 0$:

$$\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} \left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right)$$

That is, the moduli divide and the arguments subtract.

Proof

Multiply the numerator and denominator by the conjugate $\cos\theta_2 - i\sin\theta_2$ of $z_2$.

\begin{align*} \dfrac{z_1}{z_2} &= \dfrac{r_1(\cos\theta_1 + i\sin\theta_1)}{r_2(\cos\theta_2 + i\sin\theta_2)} \\ &= \dfrac{r_1}{r_2} \cdot \dfrac{(\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 - i\sin\theta_2)}{(\cos\theta_2 + i\sin\theta_2)(\cos\theta_2 - i\sin\theta_2)} \\ &= \dfrac{r_1}{r_2} \cdot \dfrac{(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 - \cos\theta_1\sin\theta_2)}{\cos^2\theta_2 + \sin^2\theta_2} \\ &= \dfrac{r_1}{r_2} \left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right) \end{align*}

The denominator becomes $\cos^2\theta_2 + \sin^2\theta_2 = 1$, and the numerator simplifies through the angle-subtraction formulas for sine and cosine.

Geometric meaning

Multiplying a complex number $z$ by $w$:

  • rotates it about the origin by $\mathrm{arg}(w)$
  • scales its distance from the origin by a factor of $|w|$

In particular, when $|w| = 1$, multiplying by $w$ is a pure rotation.

De Moivre's Theorem

Theorem: De Moivre's theorem

For any integer $n$:

$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

Proof (for $n \geq 1$, by induction)

For $n = 1$: $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$. Clearly true.

Assume it holds for $n = k$: $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$.

For $n = k+1$:

\begin{align*} (\cos\theta + i\sin\theta)^{k+1} &= (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta) \\ &= (\cos(k\theta) + i\sin(k\theta))(\cos\theta + i\sin\theta) \\ &= \cos((k+1)\theta) + i\sin((k+1)\theta) \end{align*}

The last equality uses multiplication in polar form.

For $n = 0$: $(\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$. True.

For $n < 0$: writing $n = -m$ ($m > 0$),

$$(\cos\theta + i\sin\theta)^{-m} = \dfrac{1}{(\cos\theta + i\sin\theta)^m} = \dfrac{1}{\cos(m\theta) + i\sin(m\theta)}$$

$= \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta)$.

Example: applying De Moivre's theorem

Compute $(1 + i)^8$.

From $1 + i = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$:

\begin{align*} (1 + i)^8 &= \left(\sqrt{2}\right)^8 \left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)^8 \\ &= 16 \left(\cos\dfrac{8\pi}{4} + i\sin\dfrac{8\pi}{4}\right) \\ &= 16(\cos 2\pi + i\sin 2\pi) \\ &= 16 \cdot 1 = 16 \end{align*}

nth Roots

In the complex numbers, every nonzero number has $n$ distinct $n$th roots.

Theorem: nth roots of a complex number

Let $w \neq 0$ be a complex number with $w = r(\cos\phi + i\sin\phi)$. The solutions of $z^n = w$ are:

$$z_k = \sqrt[n]{r}\left(\cos\dfrac{\phi + 2k\pi}{n} + i\sin\dfrac{\phi + 2k\pi}{n}\right) \quad (k = 0, 1, 2, \ldots, n-1)$$

That is, there are exactly $n$ solutions.

Geometric interpretation

The $n$th roots lie on the circle of radius $\sqrt[n]{r}$ centred at the origin, equally spaced (by an angle $\dfrac{2\pi}{n}$). The first root $z_0$ has argument $\dfrac{\phi}{n}$, and the remaining roots follow at steps of $\dfrac{2\pi}{n}$. When $\phi \neq 0$, $z_0$ does not lie on the real axis, and the whole regular polygon is rotated about the origin by $\dfrac{\phi}{n}$.

$w$
$\phi$
$\phi/3$
$z_0$
$z_1$
$z_2$
$\sqrt[3]{r}$
$O$
Re
Im
Figure 3: The cube roots of a complex number $w = r(\cos\phi + i\sin\phi)$ with argument $\phi$ (example: $\phi = 80^\circ$). The orange arrow shows the direction of the original $w$ (argument $\phi$). Taking the cube root trisects the argument, so $z_0$ has argument $\dfrac{\phi}{3}$; $z_1, z_2$ follow at steps of $\dfrac{2\pi}{3}$, forming an equilateral triangle on the circle of radius $\sqrt[3]{r}$. Since $\phi \neq 0$, $z_0$ is not on the real axis. Note that the orange arrow only indicates the direction of $w$; its tip is not $w$ itself. The actual point $w$ lies at radius $|w| = r$ and is not drawn in this figure.

Example: the cube roots of $w = 8(\cos 120^\circ + i\sin 120^\circ)$

Find the cube roots of $w = 8(\cos 120^\circ + i\sin 120^\circ) = -4 + 4\sqrt{3}\,i$. From the polar form $|w| = 8$, so by the $n = 3$ formula:

\begin{align*} z_k &= \sqrt[3]{8}\left(\cos\dfrac{120^\circ + 360^\circ k}{3} + i\sin\dfrac{120^\circ + 360^\circ k}{3}\right) \\ &= 2\big(\cos(40^\circ + 120^\circ k) + i\sin(40^\circ + 120^\circ k)\big) \quad (k = 0, 1, 2) \end{align*}
  • $z_0 = 2(\cos 40^\circ + i\sin 40^\circ) \approx 1.53 + 1.29\,i$
  • $z_1 = 2(\cos 160^\circ + i\sin 160^\circ) \approx -1.88 + 0.68\,i$
  • $z_2 = 2(\cos 280^\circ + i\sin 280^\circ) \approx 0.35 - 1.97\,i$

Each root lies on the circle of radius $\sqrt[3]{8} = 2$, and the argument $40^\circ = \dfrac{120^\circ}{3}$ of the first root $z_0$ shows that taking the cube root trisects the argument of $w$.

$z_0$
$z_1$
$z_2$
Figure 4: the cube roots of the example $w = 8(\cos 120^\circ + i\sin 120^\circ)$. They lie on the circle of radius $\sqrt[3]{8} = 2$, forming an equilateral triangle with $z_0$ (argument $40^\circ$) rotated in steps of $120^\circ$.

Summary

  • In the complex plane, a complex number $z = a + bi$ is drawn as the point $(a, b)$
  • In polar form $z = r(\cos\theta + i\sin\theta)$, $r$ is the modulus and $\theta$ the argument
  • Under multiplication the moduli multiply and the arguments add
  • De Moivre's theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$
  • A complex number has exactly $n$ $n$th roots, equally spaced around a circle

Frequently Asked Questions

Q1: What is the complex plane?

A: The plane in which a complex number $z = x + iy$ is drawn as the point $(x, y)$ is called the Gaussian plane (complex plane, or Argand diagram). The horizontal axis is the real part Re$(z) = x$ and the vertical axis the imaginary part Im$(z) = y$. $|z| = \sqrt{x^2 + y^2}$ is the modulus (the magnitude of the complex number) and $\arg(z) = \arctan(y/x)$ is the argument.

Q2: What is polar form?

A: Writing a complex number $z = x+iy$ as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ (with $r = |z|$, $\theta = \arg z$) is called its polar form. Multiplication becomes simple: $z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$ (the moduli multiply, the arguments add).

Q3: What is De Moivre's theorem?

A: It is the identity $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for every integer $n$. It follows naturally because the $n$th power of $e^{i\theta}$ is $e^{in\theta}$, and it is used to find the roots of $z^n = 1$ (the $n$th roots of unity).