Euler's Formula/Real Domain/Proof 2

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Theorem

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

Then:

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


Proof

This:

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

is logically equivalent to this:

$\dfrac{\cos \theta + i \sin \theta} {e^{i \theta}} = 1$

for every $\theta$.

Note that the left expression is nowhere undefined.


Taking the derivative of this:

\(\displaystyle \dfrac {\mathrm d}{\mathrm d \theta} e^{-i \theta} \left({\cos \theta + i \sin \theta}\right)\) \(=\) \(\displaystyle e^{-i\theta} \left({-\sin \theta + i \cos \theta}\right) + \left({-i e^{-i\theta} }\right) \left({\cos \theta + i \sin \theta}\right)\) Product Rule and Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta - i^2 \sin \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta + \sin \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({0}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


Thus the expression, as a function of $\theta$, is constant and so yields the same value for every $\theta$.

We know the value at at least one point, that is, when $\theta = 0$:

$\dfrac {\cos 0 + i \sin 0} {e^{0 i}} = \dfrac {1 + 0} 1 = 1$

Thus it is $1$ for every $\theta$, which verifies the above.

Hence the result.

$\blacksquare$