Euler's Formula

Theorem

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

where $${e}^{\imath \, \theta}$$ is the complex exponential function.

Thus we define the complex exponential function in terms of standard trigonometric functions.

Direct Proof 1
Take the polar form of some complex number $$z$$:


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

Differentiate with respect to $$\theta$$, using Derivative of Sine Function and Derivative of Cosine Function:


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

Take differentials:


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

From the definition of the logarithm, we have:


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

Let us apply this to our complex variable $$z$$:


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

Thus


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

Combining this with our definition of $${z}$$, we obtain:

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

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:

$$ $$ $$ $$ $$

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.

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

$$ $$

From the definitions of the sine and cosine functions:

$$ $$

we obtain from the previous equation:

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