TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02

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Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$

Statement

For every real $θ$ , $\exp (i θ) = \cos θ + i \sin θ$ .

Facts & Assumptions

Given: A real $θ$ .

[L1]

The complex exponential by its power series defines $\exp z$ as the sum of the complex series with terms $z^{n} / n!$ , where the factorial is embedded in the complex field.

[L2]

The complex exponential series converges absolutely for every complex argument states that for every $z \in C$ , the series $\sum z^{n} / n!$ converges absolutely.

[L3]

If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ states that if $0 \leq u_{n} \leq v_{n}$ eventually and $\sum v_{n}$ converges, then $\sum u_{n}$ converges.

[L4]

Every absolutely convergent complex series converges, and rearrangements preserve its sum states that every absolutely convergent complex series converges.

[L5]

For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm states that a sequence in $R^{2}$ converges exactly when each coordinate sequence converges.

[L6]

Convergent series add and scale termwise states that convergent real series add and scale termwise, with the corresponding sums.

[L7]

Grouping: if $\sum a_{k}$ converges and $(n_{j})$ is strictly increasing with $n_{0} = 0$ , the series of blocks $\sum_{k = n_{j}}^{n_{j + 1} - 1} a_{k}$ converges to the same sum states that a convergent real series may be grouped into consecutive finite blocks without changing its sum.

[L8]

Sine and cosine defined by their real power series defines sine and cosine by the real series $\sum_{k = 0}^{\infty} (- 1)^{k} θ^{2 k + 1} / (2 k + 1)!$ and $\sum_{k = 0}^{\infty} (- 1)^{k} θ^{2 k} / (2 k)!$ .

[L9]

Integer powers in the complex field defines natural complex powers by $z^{0} = 1$ and $z^{n + 1} = z^{n} z$ .

Proof

technique · direct

1.1L1L2construct

Put $a_{n} = (i θ)^{n} / n!$ , and define the parity masks by $e_{n} = a_{n}$ for even $n$ and $e_{n} = 0$ for odd $n$ , while $o_{n} = 0$ for even $n$ and $o_{n} = a_{n}$ for odd $n$ . By [L1] and [L2], $\sum a_{n}$ converges absolutely and has sum $\exp (i θ)$ .

1.2L9algebra

Induction from $i^{2} = - 1$ and the recursion in [L9] gives $i^{2 k} = (- 1)^{k}$ and $i^{2 k + 1} = i (- 1)^{k}$ for every $k \in N$ . Hence $e_{2 k} = (- 1)^{k} θ^{2 k} / (2 k)!$ , $e_{2 k + 1} = 0$ , $o_{2 k} = 0$ , and $o_{2 k + 1} = i (- 1)^{k} θ^{2 k + 1} / (2 k + 1)!$ .

2.1L3L4step 1.1

Since $0 \leq ∣ e_{n} ∣ \leq ∣ a_{n} ∣$ and $0 \leq ∣ o_{n} ∣ \leq ∣ a_{n} ∣$ , [L3] shows that both modulus series converge. Thus $\sum e_{n}$ and $\sum o_{n}$ are absolutely convergent and hence converge by [L4].

3.1L5L6step 2.1

By [L5], the two coordinate series of $e$ and $o$ converge. Since $a_{n} = e_{n} + o_{n}$ , applying [L6] in each coordinate gives $\sum_{n = 0}^{\infty} a_{n} = \sum_{n = 0}^{\infty} e_{n} + \sum_{n = 0}^{\infty} o_{n}$ .

3.2L5L7L8step 2.1step 1.2

Apply [L7] to each real coordinate series of $e$ , with consecutive blocks $[2 k, 2 k + 2)$ . By step 1.2 the real-coordinate blocks are $(- 1)^{k} θ^{2 k} / (2 k)!$ and the imaginary-coordinate blocks are $0$ . Therefore [L8] identifies $\sum e_{n} = \cos θ$ .

3.3L5L7L8step 2.1step 1.2

Apply the same coordinatewise grouping to $o$ . Its real-coordinate blocks are $0$ and its imaginary-coordinate blocks are $(- 1)^{k} θ^{2 k + 1} / (2 k + 1)!$ , so [L8] gives $\sum o_{n} = i \sin θ$ .

4.1step 1.1step 3.1step 3.2step 3.3∎

Substitute steps 3.2 and 3.3 into step 3.1 and use step 1.1: $\exp (i θ) = \cos θ + i \sin θ$ .

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential