Euler's Formula

Theorem
$${e}^{\imath \, \theta} \;=\; \cos(\theta) + \imath\,\sin(\theta)$$ (on the complex exponential function, not map theory) defines the complex exponent function in terms of standard trig functions.

Direct Proof 1
Take the polar form of some complex number

$$z \;\equiv\; \cos(\theta) + \imath\,\sin(\theta)$$

Differentiate with respect to theta.

$$\frac{dz}{d\theta} \;=\; -\sin(\theta) + \imath\,\cos(\theta) \;=\; \imath\,[\,\cos(\theta) + \imath\,\sin(\theta)\,] \;=\; \imath \, z $$

Take differentials

$${dz} \;=\; \imath \, z \; {d\theta} $$

Now in general we have defined the logarithm by

$$\ln x \;\equiv\; \int \frac{dx}{x}$$

Let us apply this to our complex variable

$$\ln z \;=\; \int \frac{dz}{z} \;=\; \int \frac{\imath \, z}{z} \, {d\theta} \;=\; \imath \, \theta $$

Thus

$${e}^{\ln z} \;=\; {e}^{\imath \, \theta}$$

And combining with our definition of $${z}$$

$${e}^{\imath \, \theta} \;=\; \cos(\theta) + \imath\,\sin(\theta)$$

QED

Direct Proof 2
If

$${e}^{\imath \, \theta} \;=\; \cos(\theta) + \imath\,\sin(\theta)$$

then

$$\frac{\cos(\theta) + \imath\,\sin(\theta)}{{e}^{\imath \, \theta}} \;=\; 1$$

for every $$\theta$$. Note that the left expression is nowhere undefined.

Taking the derivative of this:

$$\frac{d}{d\theta} e^{-\imath\theta}(\cos\theta+\imath\sin\theta)$$

$$= e^{-\imath\theta}(-\sin\theta+\imath\cos\theta) + (-\imath e^{-\imath\theta})(\cos\theta+\imath\sin\theta)$$

$$= e^{-\imath\theta}(-\sin\theta+\imath\cos\theta - \imath\cos\theta - \imath^2\sin\theta)$$

$$= e^{-\imath\theta}(-\sin\theta+\imath\cos\theta - \imath\cos\theta + \sin\theta) = e^{-\imath\theta}(0) = 0 $$

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 ($$\theta = 0$$)

$$\frac{\cos(0) + \imath\,\sin(0)}{{e}^{\imath \, 0}} \;=\; \frac{1 + 0}{1} \;=\; 1$$

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

QED

Direct Proof 3
Use the Taylor series for

$$ e^{\imath\theta} = 1 + \imath\theta + \frac{\imath^2\theta^2}{2} + \frac{\imath^3\theta^3}{3!} + \frac{\imath^4\theta^4}{4!} + \frac{\imath^5\theta^5}{5!} + \frac{\imath^6\theta^6}{6!} + \frac{\imath^7\theta^7}{7!} + \frac{\imath^8\theta^8}{8!} + \cdots$$

The equation can be simplified to

$$ e^{\imath\theta} = 1 + \imath\theta - \frac{\theta^2}{2} - \frac{\imath\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{\imath\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{\imath\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots$$

Rearranging the above equation, we obtain

$$ e^{\imath\theta} = (1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} +\cdots) + (\imath\theta - \frac{\imath\theta^3}{3!} + \frac{\imath\theta^5}{5!} - \frac{\imath\theta^7}{7!} + \cdots )$$

$$ = (1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} +\cdots) + \imath(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots ) $$

Recognizing that

$$ \sin(\theta) = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots $$

and

$$ \cos(\theta) = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots $$

We obtain from the previous equation

$$ e^{\imath\theta} = \cos(\theta) + \imath\sin(\theta) $$

QED