Logical explanation of Euler's formula This question is a about (if not proving) at least guessing the Euler's formula. 
I don't want the proof using the infinite sums.
We can guess by logic that for example that the equation $x^2+1=\sqrt{x}$ has no real solutions because $x^2=\sqrt{x}$ has 2 solutions $x=0, x=1$ but by adding 1 on the left side, we cancel these 2 solutions, so there are no solutions. 
I want to know if there is a way to guess by logic that $e^{ix}=\cos(x)+i\sin(x)$. I guess that the most important here here will be $\frac{d}{dx}e^x=e^x$. And suggestions?
 A: The power series argument, while simple, is indeed unenlightening. You can easily show that $f(\theta)=\cos\left(\theta\right)+i\sin\left(\theta\right)$ satifies both $f'\left(\theta\right)=i\cdot f\left(\theta\right)$ and $f\left(0\right)=1$, and it looks remarkably similar to one definition of $\gamma(t)=e^{\alpha t}$: the function $\gamma\,\colon\mathbb{C}\rightarrow\mathbb{C}$ that both $\left(e^{\alpha t}\right)'=\alpha e^{\alpha t}$ and $\gamma(0)=1$.
A: This isn't a proof, but an illustration as why the formula is defined in that way:
Consider $h(z) = A^z$, A > 0.  This is well-defined for z real, and you want to extend it to when z is complex.  It's enough to define it whenever z is pure imaginary.
Define $f(x) := Re(h(ix))$, $g(x) := Im(h(ix))$
So $h(ix) == f(x) + i*g(x)$
1)  $h(0) == 1$,  so $f(0) == 1$, $g(0) == 0$.
2)  The distinction between i and -i is arbitrary, so when extending real-valued functions to the complex numbers, you usually want  $h(\bar{z}) == \bar{h(z)}$, so $f(-x) == f(x)$, $g(-x) == -g(x)$
3)  $A^m * A^n == A^{m+n}$, so 
$h(i(x+y)) == h(ix) * h(iy)$
$f(x+y) + i*g(x+y) == (f(x) + i*g(x)) (f(y) + i*g(y))$
$f(x+y) + i*g(x+y) == f(x)*f(y) + i*g(x)*f(y) + i*f(x)*g(y) - g(x)*g(y)$
$f(x+y) + i*g(x+y) == (f(x)*f(y) - g(x)*g(y)) + i*(g(x)*f(y) + f(x)*g(y))$
Equating real and imaginary parts:
$f(x+y) == f(x)*f(y) - g(x)*g(y)$
$g(x+y) == f(x)*g(h) + f(y)*g(x)$
Those formulas should be very familiar, and they suggest that $A^{ix}$ behaves very much like $cos(x) + i*sin(x)$.
($cos(kx) + i*sin(kx)$) satisfies all the formulas for any $k$.  Determining which value of $k$ corresponds to a particular value of $A$ requires calculus or the equivalent.)
