Euler's Formula/Real Domain/Proof 4

Theorem

Let $\theta \in \R$ be a real number.

Then:

$e^{i \theta} = \cos \theta + i \sin \theta$

Proof

Note that the following proof, as written, only holds for real $\theta$.

Define:

$\map x \theta = e^{i \theta}$

$\map y \theta = \cos \theta + i \sin \theta$

Consider first $\theta \ge 0$.

Taking Laplace transforms:

\(\ds \map {\laptrans {\map x \theta} } s\)

\(=\)

\(\ds \map {\laptrans {e^{i \theta} } } s\)

\(\ds \)

\(=\)

\(\ds \frac 1 {s - i}\)

Laplace Transform of Exponential

\(\ds \map {\laptrans {\map y \theta} } s\)

\(=\)

\(\ds \map {\laptrans {\cos \theta + i \sin \theta} } s\)

\(\ds \)

\(=\)

\(\ds \map {\laptrans {\cos \theta} } s + i \, \map {\laptrans {\sin \theta} } s\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \frac s {s^2 + 1} + \frac i {s^2 + 1}\)

Laplace Transform of Cosine, Laplace Transform of Sine

\(\ds \)

\(=\)

\(\ds \frac {s + i} {\paren {s + i} \paren {s - i} }\)

\(\ds \)

\(=\)

\(\ds \frac 1 {s - i}\)

So $x$ and $y$ have the same Laplace transform for $\theta \ge 0$.

Now define $\tau = -\theta, \sigma = -s$, and consider $\theta < 0$ so that $\tau > 0$.

Taking Laplace transforms of $\map x \tau$ and $\map y \tau$:

\(\ds \map {\laptrans {\map x \tau} } \sigma\)

\(=\)

\(\ds \frac 1 {\sigma - i}\)

from above

\(\ds \map {\laptrans {\map y \tau} } \sigma\)

\(=\)

\(\ds \frac 1 {\sigma - i}\)

from above

So $\map x \theta$ and $\map y \theta$ have the same Laplace transforms for $\theta < 0$.

The result follows from Injectivity of Laplace Transform.

$\blacksquare$