Euler's Formula

Euler's Formula
$${e}^{\imath \, \theta} \;=\; \cos(\theta) + \imath\,\sin(\theta)$$ (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.

If one takes the derivative of this expression with respect to $$\theta$$, it is found to be zero (edit: someone do this). 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