Engineering Mathematics · Complex Analysis

Euler's Formula

Why raising e to an imaginary power makes you spin in a circle — and why that's not a coincidence.

\[ e^{i\theta} = \cos\theta + i\sin\theta \]
§ 0

What the formula actually says

Three objects that seem to have nothing to do with each other:

e — the growth number

\(e \approx 2.718...\) — the base of natural growth. Appears everywhere compounding happens: interest, populations, radioactive decay.

i — the imaginary unit

\(i = \sqrt{-1}\). A number that, when squared, gives \(-1\). Lives on a second axis perpendicular to the real number line.

cos and sin — the circle functions

\(\cos\theta\) and \(\sin\theta\) describe positions on a unit circle. Pure geometry — right triangles and angles.

θ — the angle

Measured in radians. One full revolution = \(2\pi\). Half turn = \(\pi\). Quarter turn = \(\pi/2\).

Euler's formula says: when you raise \(e\) to the power \(i\theta\), the result is a point on the unit circle at angle \(\theta\). The exponential function and the circular functions are secretly the same thing.

The one-line summary: \(e^{i\theta}\) doesn't make you grow — it makes you rotate. The imaginary unit in the exponent converts "exponential scaling" into "circular motion."

§ 1

The complex plane — where numbers have two coordinates

A real number lives on a line. A complex number lives on a plane. The horizontal axis is the real part; the vertical axis is the imaginary part.

The complex number \(a + bi\) is just the point \((a, b)\) on this plane. That's it. Nothing more mysterious than that.

Re Im 1 2 −1 −2 i −i 2 + i Re = 2 Im = 1 e^(iπ/4) π/4 unit circle (r=1) 0

The unit circle is the set of all complex numbers with magnitude 1 — all points at distance 1 from the origin. Every point on it can be written as \(\cos\theta + i\sin\theta\) for some angle \(\theta\).

\[ |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1 \]

The Pythagorean identity \(\cos^2\theta + \sin^2\theta = 1\) guarantees the magnitude is always 1. So \(\cos\theta + i\sin\theta\) walks around the unit circle as \(\theta\) increases.

§ 2

What \(e^x\) actually is — the function that grows like itself

Before we can understand \(e^{i\theta}\), we need to deeply understand \(e^x\). Here's the real definition — not a number to memorize, but a property:

\(e^x\) is the unique function that is its own derivative. That is: if \(f(x) = e^x\), then \(f'(x) = e^x\). The rate of change at every point equals the value at that point.

This means: wherever you are, you're growing at exactly the speed of your current height. Double your height, double your speed. It's pure, self-referential growth.

From this one property, we can derive the Taylor series for \(e^x\) — the infinite polynomial that equals \(e^x\) exactly everywhere:

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

Think of the Taylor series like a recipe card. \(e^x\) is so perfectly self-consistent that you can describe it completely using just the value and all derivatives at a single point — say \(x=0\). Since every derivative of \(e^x\) equals \(e^x\), and \(e^0 = 1\), every coefficient in the recipe is \(\frac{1}{n!}\). The series is the exact, infinite recipe for the function.

Why each term is x^n / n! — the derivation

A Taylor series centered at \(x=0\) is built from the rule: the coefficient of \(x^n\) is \(\frac{f^{(n)}(0)}{n!}\), where \(f^{(n)}\) means the \(n\)-th derivative.

For \(e^x\): every derivative is \(e^x\), and at \(x=0\): \(e^0 = 1\). So every coefficient is \(\frac{1}{n!}\). Therefore:

\[ e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

Since \(0! = 1\) and \(1! = 1\), this simplifies to the familiar form. The factorials grow very fast, making each successive term smaller for any finite \(x\) — guaranteeing the series converges.

The crucial point: this series works for any input — real, complex, imaginary. As long as we can plug something into the series and have it converge, we can compute \(e^{\text{that thing}}\).

§ 3

What happens when we plug in \(i\theta\)

Let's simply replace \(x\) with \(i\theta\) in the Taylor series. No tricks. Just substitution.

\[ e^{i\theta} = 1 + (i\theta) + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \cdots \]

Now we need to simplify the powers of \(i\). This is where the magic lives. Remember \(i^2 = -1\), so:

\[ i^0 = 1 \qquad i^1 = i \qquad i^2 = -1 \qquad i^3 = -i \qquad i^4 = 1 \qquad i^5 = i \quad \cdots \]

Powers of \(i\) cycle with period 4: \(1, i, -1, -i, 1, i, -1, -i, \ldots\) It's like a clock with four positions. Multiply by \(i\) once = rotate 90° counterclockwise. Do it four times = full rotation = back to start. This is already a hint that \(i\) and rotation are deeply connected.

Powers of i — each multiplication rotates 90° Re Im i⁰ = 1 i¹ = i i² = −1 i³ = −i ×i ×i ×i ×i

Now substitute the powers of \(i\) into the series. Each term \((i\theta)^n = i^n \cdot \theta^n\), so:

\[ e^{i\theta} = 1 + i\theta + \frac{i^2\,\theta^2}{2!} + \frac{i^3\,\theta^3}{3!} + \frac{i^4\,\theta^4}{4!} + \frac{i^5\,\theta^5}{5!} + \cdots \]
\[ = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \cdots \]
§ 4

Splitting into real and imaginary parts — where \(\cos\) and \(\sin\) appear

Group the terms without \(i\) (real parts) separately from the terms with \(i\) (imaginary parts):

\[ e^{i\theta} = \underbrace{\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots\right)}_{\text{real part}} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots\right)}_{\text{imaginary part}} \]

Now: what are the Taylor series for \(\cos\theta\) and \(\sin\theta\)? Let's derive them from scratch.

cos θ — Taylor series at x = 0
\[ f = \cos\theta,\quad f(0)=1 \] \[ f' = -\sin\theta,\quad f'(0)=0 \] \[ f'' = -\cos\theta,\quad f''(0)=-1 \] \[ f''' = \sin\theta,\quad f'''(0)=0 \] \[ f^{(4)} = \cos\theta,\quad f^{(4)}(0)=1 \]
sin θ — Taylor series at x = 0
\[ f = \sin\theta,\quad f(0)=0 \] \[ f' = \cos\theta,\quad f'(0)=1 \] \[ f'' = -\sin\theta,\quad f''(0)=0 \] \[ f''' = -\cos\theta,\quad f'''(0)=-1 \] \[ f^{(4)} = \sin\theta,\quad f^{(4)}(0)=0 \]

Applying the Taylor formula \(\sum \frac{f^{(n)}(0)}{n!}\theta^n\):

\[ \cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots \]
\[ \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \]

Compare these with the two halves of \(e^{i\theta}\) we found above. They are identical. Term by term, coefficient by coefficient:

Term Power of i Result Goes into...
\(\tfrac{(i\theta)^0}{0!} = 1\) i⁰ = 1 \(+1\) cos (real)
\(\tfrac{(i\theta)^1}{1!} = i\theta\) i¹ = i \(+i\theta\) sin (imaginary)
\(\tfrac{(i\theta)^2}{2!} = -\tfrac{\theta^2}{2!}\) i² = −1 \(-\tfrac{\theta^2}{2!}\) cos (real)
\(\tfrac{(i\theta)^3}{3!} = -\tfrac{i\theta^3}{3!}\) i³ = −i \(-\tfrac{i\theta^3}{3!}\) sin (imaginary)
\(\tfrac{(i\theta)^4}{4!} = +\tfrac{\theta^4}{4!}\) i⁴ = +1 \(+\tfrac{\theta^4}{4!}\) cos (real)
\(\tfrac{(i\theta)^5}{5!} = +\tfrac{i\theta^5}{5!}\) i⁵ = +i \(+\tfrac{i\theta^5}{5!}\) sin (imaginary)

The even-power terms are all real (because even powers of \(i\) are real: \(\pm 1\)) and they reassemble into \(\cos\theta\). The odd-power terms all carry \(i\) (because odd powers of \(i\) are imaginary: \(\pm i\)) and they reassemble into \(i\sin\theta\). Therefore:

Euler's Formula — derived from first principles \[ e^{i\theta} = \cos\theta + i\sin\theta \]

No hand-waving. The cycling of powers of \(i\) (period 4) routes the even terms to cosine and the odd terms to sine. That's the whole mechanism.

§ 5

Interactive — watch \(e^{i\theta}\) walk around the unit circle

Drag the slider below. Watch the point \(e^{i\theta}\) move around the circle, and see how its real part (\(\cos\theta\)) and imaginary part (\(\sin\theta\)) change.

Unit circle — e^(iθ) interactive

0.80
Re Im 1−1 i−i θ cos θ sin θ e^(iθ) e^(iθ) = cos θ = 0.70 + i·sin θ = 0.70i |e^(iθ)| = 1 always As θ increases: cos θ oscillates sin θ oscillates but e^(iθ) just spins

The key geometric insight: as \(\theta\) increases from \(0\) to \(2\pi\), the point \(e^{i\theta}\) traces exactly one full circle. The real part \(\cos\theta\) oscillates left and right. The imaginary part \(\sin\theta\) oscillates up and down. Together they trace the circle.

The magnitude \(|e^{i\theta}| = \sqrt{\cos^2\theta + \sin^2\theta} = 1\) always. No matter what \(\theta\) is. This is why the imaginary exponent produces pure rotation — it can never shrink or grow the magnitude because the Pythagorean identity forbids it.

§ 6

The famous special case: \(e^{i\pi} = -1\)

Set \(\theta = \pi\) (a half-turn, 180°).

\[ e^{i\pi} = \cos\pi + i\sin\pi = -1 + i(0) = -1 \]
Re Im 1 −1 e^(i·0) = 1 θ = 0 e^(iπ) = −1 θ = π π radians = 180° half turn cos π = −1 sin π = 0 ∴ e^(iπ) = −1

Rearrange by adding 1 to both sides:

Euler's Identity \[ e^{i\pi} + 1 = 0 \]

Five fundamental constants — \(e\), \(i\), \(\pi\), \(1\), \(0\) — in one equation. Each constant arises independently: \(e\) from calculus, \(i\) from algebra, \(\pi\) from geometry, \(1\) from counting, \(0\) from nothing. The fact that they satisfy this single clean relationship is not a trick — it's a deep structural truth about how rotation, growth, and number systems are unified.

§ 7

Why this matters — the engineering payoff

Euler's formula is not just beautiful. It's a computational superpower. Here are the three most important consequences.

1. Sinusoids become exponentials — algebra becomes trivial

Any sinusoidal signal can be written as a complex exponential:

\[ A\cos(\omega t + \phi) = \operatorname{Re}\!\left[A\,e^{i(\omega t + \phi)}\right] = \operatorname{Re}\!\left[A e^{i\phi} \cdot e^{i\omega t}\right] \]

This matters enormously in AC circuits. Instead of carrying \(\cos\) and \(\sin\) through every calculation (constantly using trig identities), you work with \(e^{i\omega t}\) — which multiplies, divides, and differentiates cleanly. The phase \(\phi\) and amplitude \(A\) bundle into the single complex number \(Ae^{i\phi}\), called a phasor.

2. Differentiation becomes multiplication

If \(f(t) = e^{i\omega t}\), then:

\[ \frac{d}{dt}e^{i\omega t} = i\omega\, e^{i\omega t} \]

Differentiating with respect to time just multiplies by \(i\omega\). That means solving differential equations for oscillating systems turns into solving algebraic equations — multiply instead of differentiate.

3. The trig identities fall out for free

From \(e^{i\theta} = \cos\theta + i\sin\theta\) alone:

\[ \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \]

And the angle addition formula becomes a single line — just multiply two exponentials:

\[ e^{i(\alpha+\beta)} = e^{i\alpha}\cdot e^{i\beta} \] \[ \Downarrow \] \[ \cos(\alpha+\beta) + i\sin(\alpha+\beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) \]

Expand the right side and match real and imaginary parts to recover the double-angle identities instantly. No memorization needed — just Euler's formula and algebra.

The deepest takeaway: Euler's formula reveals that oscillation is rotation. A sine wave in time is a shadow of circular motion in the complex plane. The Fourier transform, which decomposes any signal into frequencies, is just decomposing a function into a sum of circular rotations at different speeds — all enabled by Euler's formula.