Euler's identity

I won't tell you much but I did get the eulers identity as a tattoo on my wrist my senior year in high school. Most autistic move i ever made. You can imagine how it went for me at parties and hanging out

>>8613250

has to do with a geometric interpretation of imaginary exponents

kinda like how multiplication of complex numbers adds there angles

There's this one proof that I enjoy.

Let [math]f(x) = e^{-ix}(\cos x + i \sin x)[/math] for any [math]x \in \mathbb{R}[/math]
Then, [math]f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x) [/math] which obviously equals 0, again, for all [math]x \in \mathbb{R}[/math].
Since the derivative is 0 everywhere, we can assume the function is constant everywhere. f(0) = 1, so f(x)=1. Therefore, [math]e^{ix} = \cos x + i \sin x[/math] for all [math]x \in \mathbb{R}[/math]

>>8613267
that is really really cool. thanks.

>>8613250
Lets define an invertible function such that addition turns into multiplication: (x+y)->f(x+y)=f(x)f(y)
From this we have f(0)=1 and f(n*x)=f(x+x+...x)=f(x)^n
So this function is like an exponential function with f(x)=c^x where c=f(1)

Now lets extend this function to complex numbers z=x+i*y. f(z)=f(x+i*y)=f(x)*f(i*y)
Since we want this function to be invertible, the action of this function with complex numbers must be different so we can differentiate them. The real component scales the resulting number out radially so the imaginary component must act perpendicularly and rotate it in the complex plane. f(x+iy)=c^x*(cos(k*y)+i*sin(k*y)) where k is the rotation of f(i)
It is left as an exercise to the reader to show that k=ln(c)

>>8613267
>constant everywhere f(x)=1

so is it the radius of a complex plane unit circle?