# Analysis Incarnate

This is a quick proof of Euler’s formula without the use of the Taylor Series.

$\dpi{300}&space;\tiny&space;e^{i\phi}&space;=&space;cos&space;\phi&space;+&space;i&space;sin\phi$

Even with the imaginary number i, this equation is differentiable, so we can take the derivative of :

$\dpi{300}&space;\tiny&space;f(x)=\frac{e^{i\phi}}{cos\phi+isin\phi}$ Using the quotient rule, we end up with :

$\dpi{300}&space;\tiny&space;f\prime(x)=\frac{(cos\phi+isin\phi)ie^{i\phi}-e^{i\phi}(-sin\phi+icos\phi)}{(cos\phi+isin\phi)^{2}}$

The derivative of f(x) is $ie^{i\phi}$, and the derivative of g(x) is $e^{i\phi}$. Let’s focus on the numerator. Distributing the respective pairs, we get :

$\dpi{300}&space;\tiny&space;=e^{i\phi}(icos\phi-sin\phi+sin\phi-icos\phi)$

$\dpi{300}&space;\tiny&space;=e^{i\phi}(0)$

As you can see, the derivative of f(x)=0, and the only function whose derivative is zero is a constant. Let’s call our constant c. If we find a value for c, we know the value is the same for the entire equation, because it’s a constant. If we plug “0” into the equation that we originally derived :

$\dpi{300}&space;\tiny&space;f(0)=\frac&space;{e^{i0}}{cos0+isin0}=\frac{1}{1}$

After simplifying, we end up with (1/1), because $e^{i0}$ is one, $isin(0)$ is zero, and $cos(0)$ is one. Since we know that c is a constant, the value of f(x) remains the same.

$\dpi{300}&space;\tiny&space;f(x)=\frac&space;{e^{i\phi}}{cos\phi+isin\phi}=1$

After we simplify, we are done :)

$\dpi{300}&space;\tiny&space;e^{i\phi}=cos\phi+isin\phi$