Euler’s Identity

Let z be a complex number, so z\in\mathbb{C}.
It means we can write e^z as

    \[e^z=e^{a+ib}\]

So

    \[e^{a+ib}=e^a\cdot e^{ib}\]

Now because e^{ib}\in\mathbb{C} as well, we can write e^{ib} as

    \[e^{ib}=x+iy\]

in its turn.
If e^{ib}=x+iy then e^{-ib}=x-iy and we can see

    \[e^{ib}e^{-ib}=(x+iy)(x-iy)\]

or

    \[e^0=1=x^2+y^2\]

With e^{ib}=x+iy en x^2+y^2=1 we now know that e^{ib} in on the unity circle. It means we can write e^{ib} as polar coordinates:

    \[e^{ib}=\cos\theta+i\sin\theta\]

Now what can be said about b and \theta? Let’s assume we can differentiate in \mathbb{C} as we can in \mathbb{R}:

    \[f(b)=e^{ib}\]

so

    \[f'(b)=e^{ib}\cdot i=ie^{ib}\]

And if

    \[f(\theta)=\cos\theta+i\sin\theta\]

Then

    \[f'(\theta)=-\sin\theta+i\cos\theta=i(\cos\theta+i\sin\theta)\]

So both derivatives behave the same

    \[f'(x)=if(x)\]

And as f(0) is 1 for both functions, it means that the graphs and thus functions are the same:

    \[e^{i\theta}=\cos\theta+i\sin\theta\]

For \theta=\pi we get

    \[e^{i\pi}+1=0\]

called Euler’s identity, according to many the most beautiful formula, as it contains e, \pi, i, 1 as well as 0.