Euler's Formula

Theorem

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

where $e^{i \theta}$ 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 $z$:


 * $z \equiv \cos \theta + i \sin \theta$

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


 * $\displaystyle \frac {\mathrm d z} {\mathrm d \theta} = -\sin \theta + i \cos \theta = i \left({\cos \theta + i \sin \theta}\right) = i z$

Take differentials:


 * $\mathrm d z = i z \mathrm d \theta$

From the definition of the natural logarithm, we have:


 * $\displaystyle \ln x \equiv \int \frac {\mathrm d x} x$

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


 * $\displaystyle \ln z = \int \frac{\mathrm d z} z = \int \frac{i z} z \, \mathrm d \theta = i \theta$

Thus


 * $e^{\ln z} = e^{i \theta}$

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


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

Direct Proof 2
This:
 * $e^{i \theta} = \cos \theta + i \sin \theta$

is logically equivalent to this:
 * $\displaystyle \frac{\cos \theta + i \sin \theta} {e^{i \theta}} = 1$

for every $\theta$.

Note that the left expression is nowhere undefined.

Taking the derivative of this:

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 $\theta = 0$:


 * $\displaystyle \frac{\cos 0 + i \sin 0}{e^{i 0}} = \frac {1 + 0} 1 = 1$

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

Hence the result.

Direct Proof 3
Use the Taylor Series Expansion for Exponential Function, we have:


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

The equation can be simplified to


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

Rearranging the above eqn, we obtain

From the definitions of the sine and cosine functions:

we obtain the result we want:
 * $ e^{i \theta} = \cos \theta + i \sin \theta$