FOUNDATION · CHAPTER 1 OF 8

Complex numbers, Euler's formula and the complex exponential

This chapter builds the complex number, Euler's formula and the complex exponential e^{st} from first principles, then adds roots of unity and the geometric series. The Fourier, Laplace and z-transforms are all sums or integrals built from these objects.

Updated 12 Oct 2026

What you will be able to do

You will write any complex number in polar form and find its magnitude, angle and conjugate. You will derive Euler's formula and use it to write cos and sin as exponentials. You will read $e^{s t}$ as a steady, growing or decaying rotation. You will find the roots of unity, sum a geometric series and say when the infinite sum exists.

Before you start

Trigonometry in radians, Pythagoras and the addition formulas for $cos (α + β)$ and $sin (α + β)$ ; a calculator set to radians
The power series of $e^{x}$ , $cos x$ and $sin x$ , which converge for every real $x$
The derivatives of $e^{x}$ , $cos x$ and $sin x$ , the product rule, and $ln$
Limits of sequences, such as $a^{N} \to 0$ as $N \to \infty$ when $∣ a ∣ < 1$
The imaginary unit $j$ , with $j^{2} = - 1$ ( $j$ , never $i$ , in this subject)

Complex numbers as points and rotations

A complex number $z = a + j b$ is a point in a plane: $a$ is the real part and $b$ the imaginary part, both real. The same point can be named by its distance from the origin and its angle. Signals use this polar description, because in it multiplication is simple: the next section shows that magnitudes multiply and angles add.

MAGNITUDE AND ANGLE

z = a + j b = r cos θ + j r sin θ, r = ∣ z ∣ = a^{2} + b^{2}, cos θ = \frac{a}{r}, sin θ = \frac{b}{r}

The angle needs $z \neq = 0$ and is fixed only up to whole turns of $2 π$ . $r$ is the magnitude (same unit as $a$ and $b$ ), the hypotenuse of the right triangle with sides $∣ a ∣$ , $∣ b ∣$ . $θ$ is in radians, anticlockwise from the positive real axis. For $a \neq = 0$ , $tan θ = b / a$ , but tan alone does not fix the quadrant.

CONJUGATE

z^{*} = a - j b, z z^{*} = a^{2} + b^{2} = ∣ z ∣^{2}, z + z^{*} = 2 a, z - z^{*} = 2 j b

The conjugate $z^{*}$ is the reflection of $z$ in the real axis. The product is $(a + j b) (a - j b) = a^{2} - j^{2} b^{2} = a^{2} + b^{2}$ , using $j^{2} = - 1$ . So $z z^{*}$ is real and never negative. To divide by a non-zero $z$ , multiply top and bottom by $z^{*}$ .

WATCH OUT FOR

$arctan (b / a)$ gives the same angle for $(- 3, 4)$ and $(3, - 4)$ . Test: name the quadrant from the signs of $a$ and $b$ , then check that $r cos θ$ and $r sin θ$ give back $a$ and $b$ .

WORKED EXAMPLE

Write $z = - 3 + j 4$ in polar form. Find $∣ z ∣$ , the angle in radians and $z z^{*}$ .

Solution

Here $a = - 3 < 0$ and $b = 4 > 0$ , so the point is upper left and the angle lies between $π /2$ and $π$ .
By Pythagoras, $∣ z ∣ = 9 + 16 = 5$ .
A calculator gives $arctan (4/ (- 3)) = - 0.9273$ rad. That is correct for tan, but arctan only returns angles between $- π /2$ and $π /2$ , so it points lower right, the opposite direction.
Tan repeats every $π$ , so adding $π$ keeps the tan and turns to the upper left: $θ = - 0.9273 + 3.1416 = 2.214$ rad ( $126. 9^{\circ}$ ).
Check: $5 cos 2.214 = - 3.00$ and $5 sin 2.214 = 4.00$ , so the angle is right.
With $z^{*} = - 3 - j 4$ , $z z^{*} = 9 + 16 = 25 = ∣ z ∣^{2}$ , real as the conjugate formula says.

Answer: $z = 5 (cos 2.214 + j sin 2.214)$ , $∣ z ∣ = 5$ , $θ \approx 2.214$ rad, $z z^{*} = 25$ .

Euler's formula and the complex exponential

For complex $x$ , $e^{x}$ is defined by the same power series as for real $x$ . Put $x = j θ$ with $θ$ real. The even powers of $j$ are real and the odd powers carry one factor $j$ . Split the first $2 M + 2$ terms, a finite sum, into even and odd powers. As $M$ grows, the two parts tend to the series of $cos θ$ and $j$ times the series of $sin θ$ .

EULER'S FORMULA

e^{j θ} = k = 0 \sum \infty \frac{( j θ ) ^{k}}{k !} = m = 0 \sum \infty \frac{( - 1 ) ^{m} θ ^{2 m}}{( 2 m )!} + j m = 0 \sum \infty \frac{( - 1 ) ^{m} θ ^{2 m + 1}}{( 2 m + 1 )!} = cos θ + j sin θ

Every real $θ$ , in radians. The even terms are $k = 2 m$ and the odd terms $k = 2 m + 1$ , with $j^{2 m} = (j^{2})^{m} = (- 1)^{m}$ and $j^{2 m + 1} = j (- 1)^{m}$ . Since $cos^{2} θ + sin^{2} θ = 1$ , $∣ e^{j θ} ∣ = 1$ : the point lies on the unit circle (radius 1, centred on the origin) at angle $θ$ .

So any non-zero $z = r (cos θ + j sin θ)$ is $r e^{j θ}$ . By Euler's formula, $e^{j α} e^{j β} = (cos α + j sin α) (cos β + j sin β)$ . Multiply out with $j^{2} = - 1$ and use the addition formulas: the real part is $cos (α + β)$ and the imaginary part $sin (α + β)$ . So $e^{j α} e^{j β} = e^{j (α + β)}$ : multiplying complex numbers multiplies magnitudes and adds angles, and a factor of magnitude 1 only turns a point.

COSINE AND SINE AS SUMS OF EXPONENTIALS

e^{- j θ} = cos θ - j sin θ \Rightarrow cos θ = \frac{e ^{j θ} + e ^{- j θ}}{2}, sin θ = \frac{e ^{j θ} - e ^{- j θ}}{2 j}

Real $θ$ . The first equality is Euler's formula at $- θ$ , since cos is even and sin is odd. Add the two Euler equations and divide by 2 for cos; subtract them and divide by $2 j$ for sin. So $cos θ$ is the average of two unit points turning in opposite directions, a conjugate pair.

Now give the exponent a real part. Write $s = σ + j ω$ with $σ$ , $ω$ real, and define $e^{s t} = e^{σ t} e^{j ω t}$ : magnitude $e^{σ t}$ , angle $ω t$ . It is a point turning at $ω$ rad/s whose distance from the origin decays if $σ < 0$ , stays 1 if $σ = 0$ and grows if $σ > 0$ . With $ω = 0$ it is a plain real exponential.

THE COMPLEX EXPONENTIAL

e^{s t} = e^{σ t} (cos ω t + j sin ω t), \frac{d}{d t} e^{s t} = s e^{s t}, e^{s (t - t_{0})} = e^{- s t_{0}} e^{s t}

$σ$ in s $^{- 1}$ , $ω$ in rad/s, $t$ and the delay $t_{0}$ in s. Derivative, taking real and imaginary parts separately: $\frac{d}{d t} e^{j ω t} = - ω sin ω t + j ω cos ω t = j ω e^{j ω t}$ , so the product rule gives $σ e^{s t} + j ω e^{s t}$ . Delay: split $e^{σ (t - t_{0})}$ and $e^{j ω (t - t_{0})}$ by the exponent laws.

A signal in continuous time is written $x (t)$ , with round brackets. A system is a rule that turns an input signal $x (t)$ into an output $y (t)$ . If the output is built from the input by adding, scaling by constants, differentiating and delaying, each step turns $e^{s t}$ into a constant times $e^{s t}$ . The output is then $H (s) e^{s t}$ , where $H (s)$ , the system function, depends on $s$ but not on $t$ . That makes $e^{s t}$ the standard test signal.

WATCH OUT FOR

$ω$ in $e^{j ω t}$ is in rad/s, not hertz. Test: one full turn needs $ω T = 2 π$ , so the period is $T = 2 π / ω$ and the frequency is $f = ω /2 π$ .

WORKED EXAMPLE

Let $x (t) = e^{s t}$ with $s = - 2 + j 10$ s $^{- 1}$ and $t$ in seconds. Find its magnitude, angle, real part and imaginary part at $t = 0.5$ s, and the time at which its magnitude falls to 1% of its value at $t = 0$ .

Solution

Read off $σ = - 2$ s $^{- 1}$ and $ω = 10$ rad/s.
The magnitude is $e^{σ t} = e^{- 2 \times 0.5} = e^{- 1} = 0.368$ . Only $σ$ affects it.
The angle is $ω t = 10 \times 0.5 = 5$ rad.
In radians, the real part is $0.368 cos 5 = 0.368 \times 0.284 = 0.104$ and the imaginary part $0.368 sin 5 = 0.368 \times (- 0.959) = - 0.353$ .
The magnitude at $t = 0$ is $e^{0} = 1$ . Set $e^{- 2 t} = 0.01$ and take $ln$ : $t = ln 100/2 = 2.30$ s. $ω$ plays no part, because rotation does not change magnitude.

Answer: At $t = 0.5$ s: magnitude $0.368$ , angle $5$ rad, $x \approx 0.104 - j 0.353$ . The magnitude reaches 1% at $t \approx 2.30$ s.

Roots of unity and the geometric series

Solve $z^{N} = 1$ for a whole number $N \geq 1$ . Write $z = r e^{j θ}$ with $r > 0$ , so $z^{N} = r^{N} e^{j N θ}$ , because angles add. Since $∣ e^{j N θ} ∣ = 1$ , magnitudes give $r^{N} = 1$ , so $r = 1$ . By Euler's formula, $cos N θ + j sin N θ = 1$ needs $cos N θ = 1$ and $sin N θ = 0$ , which holds exactly when $N θ = 2 π k$ for an integer $k$ (positive, negative or zero).

ROOTS OF UNITY

z_{k} = e^{j 2 π k / N}, k = 0, 1, \dots, N - 1, W_{N} = e^{- j 2 π / N}, z_{k} = W_{N}^{- k}

$N \geq 1$ a whole number, $k$ an integer. The $N$ roots are distinct, on the unit circle, $2 π / N$ rad apart; $k$ and $k + N$ give the same point, so $k$ stops at $N - 1$ . The discrete Fourier transform (DFT) of $N$ samples $x [n]$ (square brackets for discrete time) is $X [k] = \sum_{n = 0}^{N - 1} x [n] W_{N}^{k n}$ .

FINITE GEOMETRIC SERIES

S_{N} = 1 + r + \dots + r^{N - 1}, r S_{N} = r + \dots + r^{N} \Rightarrow (1 - r) S_{N} = 1 - r^{N} \Rightarrow S_{N} = k = 0 \sum N - 1 r^{k} = \frac{1 - r ^{N}}{1 - r}

$r$ is any complex number except 1; $N \geq 1$ is a whole number. Subtracting $r S_{N}$ from $S_{N}$ cancels every middle term; then divide by $1 - r \neq = 0$ . If $r = 1$ , each term is 1 and $S_{N} = N$ .

INFINITE GEOMETRIC SERIES

k = 0 \sum \infty r^{k} k = m \sum \infty r^{k} = N \to \infty lim \frac{1 - r ^{N}}{1 - r} = \frac{1}{1 - r} = r^{m} k = 0 \sum \infty r^{k} = \frac{r ^{m}}{1 - r}

Only for $∣ r ∣ < 1$ ; $m \geq 0$ a whole number. $∣ r^{N} ∣ = ∣ r ∣^{N}$ tends to 0 exactly when $∣ r ∣ < 1$ . If $∣ r ∣ \geq 1$ , each step $S_{N + 1} - S_{N} = r^{N}$ has size at least 1, but partial sums tending to a limit need steps tending to 0, so the sum does not exist. The second line takes $r^{m}$ out of every term.

WATCH OUT FOR

Using $1/ (1 - r)$ without checking $r$ . Test: compute $∣ r ∣$ first. $r = 0.5 j$ passes ( $∣ r ∣ = 0.5$ ); $r = - 1$ and $r = e^{j 2}$ fail, since $∣ e^{j θ} ∣ = 1$ for every real $θ$ .

WORKED EXAMPLE

(a) Add the four fourth roots of unity using the finite series. (b) Evaluate $\sum_{k = 0}^{\infty} (0.5 j)^{k}$ .

Solution

(a) The roots are $1, j, - 1, - j$ , that is $r^{k}$ for $k = 0, \dots, 3$ with $r = e^{j π /2} = j$ .
Here $r \neq = 1$ , so the finite series applies: $S_{4} = (1 - j^{4}) / (1 - j) = 0$ , because $j^{4} = (j^{2})^{2} = 1$ . Directly: $1 + j - 1 - j = 0$ .
For any $N \geq 2$ , take $r = e^{j 2 π / N}$ . Its angle lies strictly between 0 and $2 π$ , so $r \neq = 1$ , while $r^{N} = e^{j 2 π} = 1$ . The same steps give 0.
(b) The ratio is $r = 0.5 j$ , with $∣ r ∣ = 0.5 < 1$ , so the series converges to $1/ (1 - 0.5 j)$ .
Multiply top and bottom by the conjugate $1 + 0.5 j$ : the bottom becomes $1 + 0.25 = 1.25$ , so the sum is $0.8 + j 0.4$ . Check: the first four terms add to $0.75 + j 0.375$ , already close.

Answer: (a) $0$ , and for any $N \geq 2$ the $N$ roots of unity add to 0. (b) $0.8 + j 0.4$ .

Check yourself

CHECK YOURSELF 1

What is the principal angle (in the range $- π < θ \leq π$ ) of $z = - 1 - j$ ?

(A)
$π /4$
(B)
$- π /4$
(C)
$3 π /4$
(D)
$- 3 π /4$

Answer and solution

Answer: (D)

The point $(- 1, - 1)$ is in the lower left quadrant, so the angle lies between $- π$ and $- π /2$ .
$arctan (1) = π /4$ is the angle between the point and the negative real axis.
So $θ = - π + π /4 = - 3 π /4$ . Check: $2 cos (- 3 π /4) = - 1$ and $2 sin (- 3 π /4) = - 1$ .

CHECK YOURSELF 2

A signal is $x (t) = e^{s t}$ with $s = - 0.5 + j 3$ s $^{- 1}$ and $t$ in seconds. Find $∣ x (t) ∣$ at $t = 2$ s. Give the value to three decimal places.

Answer and solution

Answer: $0.368$ (accepted range $\pm 0.005$ )

Only the real part $σ = - 0.5$ s $^{- 1}$ affects the magnitude, since $∣ e^{j 3 t} ∣ = 1$ .
$∣ x (2) ∣ = e^{- 0.5 \times 2} = e^{- 1}$ .
$e^{- 1} \approx 0.368$ .

CHECK YOURSELF 3One or more options are correct

For real $θ$ , which statements are correct?

(A)
$cos θ = \frac{e ^{j θ} + e ^{- j θ}}{2}$
(B)
$sin θ = \frac{e ^{j θ} + e ^{- j θ}}{2 j}$
(C)
$∣ e^{j θ} ∣ = 1$
(D)
$e^{j θ} e^{- j θ} = 1$

Answer and solution

Answer: (A), (C), (D)

A is correct: add the Euler equations for $e^{j θ}$ and $e^{- j θ}$ and divide by 2.
B is wrong: for sin the exponentials are subtracted, giving $(e^{j θ} - e^{- j θ}) / (2 j)$ .
C is correct: $∣ e^{j θ} ∣^{2} = cos^{2} θ + sin^{2} θ = 1$ .
D is correct: $e^{j α} e^{j β} = e^{j (α + β)}$ with $α = θ$ and $β = - θ$ gives $e^{j 0} = cos 0 + j sin 0 = 1$ .

CHECK YOURSELF 4

Let $W_{8} = e^{- j 2 π /8}$ . Find the real part of $W_{8}^{3}$ . Give the value to three decimal places.

Answer and solution

Answer: $- 0.707$ (accepted range $\pm 0.005$ )

Angles add, so $W_{8}^{3} = e^{- j 3 \times 2 π /8} = e^{- j 3 π /4}$ .
By Euler's formula the real part is $cos (- 3 π /4) = cos (3 π /4)$ , since cos is even.
$cos (3 π /4) = - 1/ 2 \approx - 0.707$ .

Complex numbers, Euler's formula and the complex exponential

What you will be able to do

Before you start

Complex numbers as points and rotations

Euler's formula and the complex exponential

Roots of unity and the geometric series

Check yourself

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