Euler's Identity | Graduate Proofs | Mathematics

Two proofs of e^(iπ) + 1 = 0: from the power series for exp, sin and cos, and from the circle group.

1 Statement

Theorem 1.1 (Euler’s identity).

\[ e^{i\pi } + 1 = 0. \]

2 Setting

Everything is built from power series, so no geometric intuition about angles is assumed. Section 5 gives a second proof using group theory.

Definition 2.1 (Exponential, sine and cosine). For \(z \in \C \), define

\begin{equation}\tag{2.1} \exp (z) = \sum _{n=0}^{\infty } \frac {z^n}{n!}, \qquad \cos (z) = \sum _{n=0}^{\infty } \frac {(-1)^n z^{2n}}{(2n)!}, \qquad \sin (z) = \sum _{n=0}^{\infty } \frac {(-1)^n z^{2n+1}}{(2n+1)!}, \end{equation}

and write \(e^{z} = \exp (z)\).

Definition 2.2 (The number \(\pi \)). \(\pi \) is twice the smallest positive real zero of \(\cos \).

3 Idea of the proof

Substitute \(i\theta \) into the series for \(\exp \). The powers of \(i\) cycle through \(1, i, -1, -i\), so the even terms rebuild \(\cos \theta \) and the odd terms rebuild \(i \sin \theta \). Then evaluate at \(\theta = \pi \).

4 Proof

Lemma 4.1 (Absolute convergence). Each series in (2.1) converges absolutely for every \(z \in \C \).

Proof. For \(\exp \), the ratio of consecutive terms of \(\sum \abs {z}^n / n!\) is \(\abs {z} / (n+1) \to 0\), so the series converges by the ratio test. The series for \(\cos \) and \(\sin \) are bounded term by term by this one.

Lemma 4.2 (Euler’s formula). For every \(\theta \in \R \), \(e^{i\theta } = \cos \theta + i \sin \theta \).

Proof. Since \(i^{2k} = (-1)^k\) and \(i^{2k+1} = (-1)^k i\), by Lemma 4.1 we may split the series into even and odd terms:

\begin{align*} e^{i\theta } &= \sum _{k=0}^{\infty } \frac {(i\theta )^{2k}}{(2k)!} + \sum _{k=0}^{\infty } \frac {(i\theta )^{2k+1}}{(2k+1)!} \\ &= \sum _{k=0}^{\infty } \frac {(-1)^k \theta ^{2k}}{(2k)!} + i \sum _{k=0}^{\infty } \frac {(-1)^k \theta ^{2k+1}}{(2k+1)!} = \cos \theta + i \sin \theta . \qedhere \end{align*}

Lemma 4.3 (Values at \(\pi \)). \(\cos \pi = -1\) and \(\sin \pi = 0\).

Proof. Differentiating (2.1) term by term gives \(\sin ' = \cos \) and \(\cos ' = -\sin \), so \(\frac {\dd }{\dd x}(\sin ^2 x + \cos ^2 x) = 0\) and \(\sin ^2 + \cos ^2 = 1\).

By Definition 2.2, \(\cos (\pi /2) = 0\), so \(\sin (\pi /2) = \pm 1\). Since \(\sin ' = \cos > 0\) on \([0, \pi /2)\) and \(\sin 0 = 0\), \(\sin (\pi /2) = 1\). The addition formulas, which follow from \(e^{a+b} = e^a e^b\) and Lemma 4.2, then give

\[ \cos \pi = \cos ^2\tfrac {\pi }{2} - \sin ^2\tfrac {\pi }{2} = -1, \qquad \sin \pi = 2 \sin \tfrac {\pi }{2} \cos \tfrac {\pi }{2} = 0. \qedhere \]

Proof of Theorem 1.1. By Lemmas 4.2 and 4.3,

\[ e^{i\pi } = \cos \pi + i \sin \pi = -1, \]

so \(e^{i\pi } + 1 = 0\).

5 A second proof: the circle group

The first proof computes \(\cos \pi \) and \(\sin \pi \) through the addition formulas. The second avoids them: it needs only the value at \(\pi /2\) and the fact that \(\exp \) turns addition into multiplication.

Lemma 5.1 (Homomorphism). \(\exp (z + w) = \exp (z)\exp (w)\) for all \(z, w \in \C \), and \(\exp (0) = 1\).

Proof. The series for \(\exp (z)\) and \(\exp (w)\) converge absolutely by Lemma 4.1, so their Cauchy product converges to \(\exp (z)\exp (w)\). Its \(n\)th term is

\[ \sum _{k=0}^{n} \frac {z^k}{k!} \frac {w^{n-k}}{(n-k)!} = \frac {1}{n!} \sum _{k=0}^{n} \binom {n}{k} z^k w^{n-k} = \frac {(z + w)^n}{n!} \]

by the binomial theorem, which is the \(n\)th term of \(\exp (z + w)\).

Let \(\mathbb {T} = \set {w \in \C : \abs {w} = 1}\). It is a group under multiplication, called the circle group.

Proposition 5.2 (Into the circle). The map \(E(\theta ) = e^{i\theta }\) is a group homomorphism \(E \colon (\R , +) \to (\mathbb {T}, \cdot )\).

Proof. The series for \(\exp \) has real coefficients, so \(\overline {e^{i\theta }} = e^{-i\theta }\). By Lemma 5.1,

\[ \abs {e^{i\theta }}^2 = e^{i\theta }\, e^{-i\theta } = e^{0} = 1, \]

so \(E\) takes values in \(\mathbb {T}\). It is a homomorphism by Lemma 5.1.

Lemma 5.3 (A quarter turn). \(E(\pi /2) = i\).

Proof. By Lemma 4.2 and Definition 2.2, \(E(\pi /2) = \cos \frac \pi 2 + i \sin \frac \pi 2 = i \sin \frac \pi 2\). This has modulus \(1\) by Proposition 5.2, so \(\sin \frac \pi 2 = \pm 1\). On \([0, \pi /2)\) we have \(\sin ' = \cos > 0\) and \(\sin 0 = 0\), so \(\sin \frac \pi 2 > 0\) and \(E(\pi /2) = i\).

Second proof of Theorem 1.1. \(E\) is a homomorphism, so

\[ e^{i\pi } = E\bigl (\tfrac \pi 2 + \tfrac \pi 2\bigr ) = E\bigl (\tfrac \pi 2\bigr )^2 = i^2 = -1. \qedhere \]

Remark 5.4 (The group-theoretic meaning). More is true [2, Prologue]: \(E\) is surjective with kernel \(2\pi \Z \), so it induces an isomorphism of groups

\[ \R / 2\pi \Z \;\cong \; \mathbb {T}, \qquad \theta + 2\pi \Z \mapsto e^{i\theta }. \]

Each side has exactly one element of order \(2\): the class of \(\pi \) on the left, and \(-1\) on the right, since \(w^2 = 1\) forces \(w = \pm 1\). An isomorphism must match them. In this language, Euler’s identity says that half a turn is \(-1\).

6 Remarks

Remark 6.1. The identity links five constants (\(0\), \(1\), \(e\), \(i\), \(\pi \)) using addition, multiplication and exponentiation exactly once each. The proof shows it is less mysterious than it looks: it is Lemma 4.2 at a single point.

Remark 6.2. Two facts are used without proof here: that power series can be differentiated term by term inside their radius of convergence, and that \(\cos \) has a positive zero (true since \(\cos 0 = 1\) and \(\cos 2 < 0\) from the alternating series).

References

[1]: W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw–Hill, 1976, ch. 8.
[2]: W. Rudin, Real and Complex Analysis, 3rd ed., McGraw–Hill, New York, 1987.