Is this a valid proof for eulers formula? I am wondering whether this proof is a valid proof of Eulers formula: $e^{ix}=i\sin(x)+\cos(x)$
$$\frac{d}{dx}e^{ix} = i(e^{ix})$$
$$\frac{d}{dx}(i\sin(x)+\cos(x)) = i\cos(x)-\sin(x) = i(i\sin(x)+cos(x))$$
Therefore it follows that for both $f(x)=e^{ix}$ and $f(x)=i\sin(x)+\cos(x)$, the following statement is true: 
$$\frac{d}{dx}f(x)=i(f(x))$$
And therefore since they both follow this property, they are equivalent. 
I am very well aware of the actual proof involving taylor series, and my instinct is telling me that there is no justification for that final statement thus making it wrong, so now I am wondering, is there anything that can be done to this to make it a valid proof of Euler's formula?
 A: 1) why is $\frac{d}{dx}(e^{ix})=ie^{ix})$. If you want to prove this, then you need to know the definition of exponentiation of imaginary numbers. You also need to know the definition of differentiation of functions from $\mathbb{R}$ to $\mathbb{C}$, but this is an easier problem.
2) Why do two functions $f_1,f_2:\mathbb{R}\rightarrow\{z\in\mathbb{C}|Re(z)=0\}$ that satisfy $f_1'=f_2'$ must be equal ?
A: I like using Wolfram's proof:
$$z\equiv\cos(\theta)+i\sin(\theta)\tag0$$
$$\begin{align}dz=&-\sin(\theta)+i\cos(\theta)\ d\theta\\=&i(\cos(\theta)+i\sin(\theta))\ d\theta\\=&iz\ d\theta\\\int\frac{dz}z=&\int i\ d\theta\end{align}$$
$$\ln(z)=i\theta+c$$
$$z=e^{i\theta+c}\equiv\cos(\theta)+i\sin(\theta)\tag0$$
Plugging in $\theta=0$ yields $c=0$, hence
$$\bbox[5px,border:2px solid #C0A000]{e^{i\theta}=\cos(\theta)+i\sin(\theta)}$$
A: You at least need to verify that the two functions satisfy a common initial condition, but other than that it looks pretty good. The real question is whether a complex-valued function is uniquely determined by its initial conditions and the complex-coefficient differential equation that it satisfies. If that is true, then it looks like you have a valid proof.
A: here is a similar one with a little less complex numbers,
$$f(x)=e^{ix}$$
$$g(x)=\cos x + i\sin x$$
Now one can show that $f(x)$ and $g(x)$ are solutions of $y''+ y=0$, whıch ıs a lınear homogeneous ode.
We proceed by ınvestıgatıng the lınear ındependence of $f$ and $g$. Thus we compute the Wronskıan and ıt ıs just zero.
$$W(f(x),g(x))= f(x)g'(x)-f'(x)g(x)=0$$
$$\forall x\space W=0$$
Now we are sure that functıons $f$ and $g$ are lınearly dependent. We can wrıte
$$f(x)=Cg(x) \text{ for some constant C.}$$
Now one can show that $f(0)=1$ and $g(0)=1$.
Then we conclude that $C=1$ and thus $f(x)=g(x).$
