---
tag: X4XV12DX48
kind: proof
name: Proof of Euler's formula by power series
msc: [33B10]
proves: 04R3CY41VJ
uses: [054VXHMCA3, 0VARVQ9ZW1, 07KBQCG9KV, 0KW2X0TFBX, 0RAM4E07NR, 0734RSVF2V, 0VHAVXW08V]
sources:
  - url: https://en.wikipedia.org/wiki/Euler%27s_formula
    locator: Using power series
---

## Proof

We compare partial sums of the power series.

1. A sequence $(u_N)$ of complex numbers converges to $u$ when $|u_N - u| \to 0$, with $|\cdot|$ the [[modulus|0KW2X0TFBX]].
For each complex $v$, $|\operatorname{Re} v| \le |v|$, $|\operatorname{Im} v| \le |v|$ and $|v| \le |\operatorname{Re} v| + |\operatorname{Im} v|$.
These follow by squaring both sides, since $|v|^2 = (\operatorname{Re} v)^2 + (\operatorname{Im} v)^2$.
So $u_N \to u$ holds exactly when $\operatorname{Re} u_N \to \operatorname{Re} u$ and $\operatorname{Im} u_N \to \operatorname{Im} u$.
2. By step 1, rules for real limits carry over to complex sequences, part by part.
So [[limits of sums|0RAM4E07NR]] are sums of limits, and the same holds for [[products|0734RSVF2V]] with a constant.
Also, [[subsequences|0VHAVXW08V]] of a convergent sequence converge to the same limit.
3. For complex $w$ and $M \ge 0$, let $E_M(w) = \sum_{k=0}^{M} w^k / k!$.
By the definition of the [[complex exponential|054VXHMCA3]], $E_M(w) \to e^w$ as $M \to \infty$.
4. Since [[$i^2 = -1$|07KBQCG9KV]], we get $i^{2n} = (-1)^n$ and $i^{2n+1} = (-1)^n i$ for each $n \ge 0$.
Also $(i z)^k = i^k z^k$ and $(-i z)^k = (-1)^k i^k z^k$.
5. Fix $z \in \C$. For $N \ge 0$ let $C_N = \sum_{n=0}^{N} (-1)^n \dfrac{z^{2n}}{(2n)!}$ and $S_N = \sum_{n=0}^{N} (-1)^n \dfrac{z^{2n+1}}{(2n+1)!}$.
These are the partial sums of the power series of $\cos z$ and $\sin z$. Let $M = 2N + 1$.
6. By step 4, $E_M(i z) + E_M(-i z) = \sum_{k=0}^{M} \bigl(1 + (-1)^k\bigr) i^k z^k / k!$.
The terms with odd $k$ vanish, and the term with $k = 2n$ is $2 (-1)^n z^{2n} / (2n)!$. So the sum is $2 C_N$.
7. In the same way, $E_M(i z) - E_M(-i z) = \sum_{k=0}^{M} \bigl(1 - (-1)^k\bigr) i^k z^k / k!$.
The terms with even $k$ vanish, and the term with $k = 2n + 1$ is $2 (-1)^n i z^{2n+1} / (2n+1)!$. So the sum is $2 i S_N$.
8. As $N \to \infty$, the sequences $E_{2N+1}(i z)$ and $E_{2N+1}(-i z)$ are subsequences of the sequences in step 3.
By step 2, they converge to $e^{i z}$ and $e^{-i z}$.
9. By steps 6 to 8 and step 2, $C_N \to \dfrac{e^{i z} + e^{-i z}}{2}$ and $S_N \to \dfrac{e^{i z} - e^{-i z}}{2 i}$.
So the power series of $\cos z$ and $\sin z$ converge, and these limits are $\cos z$ and $\sin z$.
10. Therefore
$$\cos z + i \sin z = \frac{e^{i z} + e^{-i z}}{2} + \frac{e^{i z} - e^{-i z}}{2} = e^{i z}.$$

## Idea

The powers of $i$ repeat with period four: $1, i, -1, -i$.
So in the series for $e^{i z}$ the even terms are real multiples of $z^{2n}$ and give $\cos z$, while the odd terms give $i \sin z$.
Adding and subtracting the series for $e^{-iz}$ separates the two halves cleanly, so no rearrangement of an infinite series is needed.