TheoremProved

Euler's formula

Statement

For any real $\theta$ , $e^{i\theta} = \cos\theta + i\sin\theta$ .

Why is it true?

Differentiating $\cos\theta+i\sin\theta$ with respect to $\theta$ gives $-\sin\theta+i\cos\theta = i(\cos\theta+i\sin\theta)$ : the function grows in the direction $i$ times itself, at every point rotated $90°$ from where it is — exactly the behaviour of $e^{i\theta}$ . Both trace the same unit circle at unit angular speed, which is why they are the same function. Setting $\theta=\pi$ gives Euler's identity $e^{i\pi}+1=0$ .

Proof sketch

Substitute $x=i\theta$ into the Taylor series $e^x=\sum_{k=0}^\infty x^k/k!$ and separate even and odd powers of $i$ : the even-power terms reassemble into $\sum (-1)^m\theta^{2m}/(2m)! = \cos\theta$ and the odd-power terms into $i\sum (-1)^m\theta^{2m+1}/(2m+1)! = i\sin\theta$ , using $i^2=-1$ .

Proved by

proved 1748Leonhard Euler

Topics that use this theorem

Complex numbers Trigonometry

Related theorems

De Moivre's formula Fundamental theorem of algebra

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

John Stillwell (2010). Mathematics and Its History