How to prove Eulers formula using the definition of derivative?

Question

A lot of proofs i have seen where people have proven Eulers formula, the first step have often been deriving $y=e^{ix}$ with the help of the chain rule and written it like $dy/dx=ie^{ix}$. According to all derivations I have seen for the chain rule formula it is assumed that the difference in the denominator $g(x)-g(a)\to 0$ (see image), where $x \to a$. For example this proof:

Here it is assumed that $(g(x)-g(a))\to 0$. If $g(x)$ for example was $e^{ix}$, the difference between $g(x)$ and $g(a)$ would go toward imaginary zero and therefore would the denominator also go towards imaginary zero which does not correspond to the definition of the derivative of any function. How is the chain rule formula derived for functions involving imaginary numbers?

Answer 1

Every real number could be considered as a complex number with its imaginary part being $0$

Thus the real zero and the complex zero are basically the same when considered as complex numbers.

Answer 2

Your question mixes up a lot of things. You start with two differentiable functions $f : U \to V, g : V \to W$ and state the chain rule in the form $(f \circ g)'(a) = f'(g(a)) \cdot g'(a)$. This makes only sense if $U, V, W$ are open subsets of $\mathbb{R}$ and $f,g$ are differentiable functions of a real variable or if $U, V, W$ are open subsets of $\mathbb{C}$ and $f,g$ are differentiable functions of a complex variable.

You have a function $g : \mathbb{R} \to \mathbb{C}, g(x) = ix$ and a function $f : \mathbb{C} \to \mathbb{C}, f(z) = e^{iz}$. The above chain rule is not applicable in this situation.

However, you can use the chain rule of multivariable calculus. Here you deal with functions of several real variables $\varphi : U \to V$ where $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^m$ are open. In this case you may regard the derivative $D\varphi$ at $a \in U$ as the Jacobian matrix of $\varphi$ at $a$. In the chain rule of multivariable calculus you do not multiply (real or complex) numbers, but matrices.

In your case you must regard $f, g$ as functions $g : \mathbb{R} \to \mathbb{R}^2, f : \mathbb{R}^2 \to \mathbb{R}^2$ and apply the matrix chain rule. But now you can prove the following little theorem ("mixed chain rule"):

Consider two functions $g : U \to V, f : V \to W$ with $U \subset \mathbb{R}$ open and $V, W \subset \mathbb{R}^2 = \mathbb{C}$ open. Assume that $g$ is differentiable in the real variable sense and $f$ is complex differentiable (which implies that is differentiable in the real variable sense).

Write $g(x) = (g_1(x),g_2(x)) \in \mathbb{R}^2 = \mathbb{C}$ with the real-valued coordinate function $g_i$ and define $g'(x) = (g'_1(x),g'_2(x)) \in \mathbb{R}^2 = \mathbb{C}$. The $g'_i(x)$ are the ordinary derivatives of real functions. Similarly you interpret $(f \circ g)'(x) \in \mathbb{R}^2 = \mathbb{C}$. Then $$(f \circ g)'(a) = f'(g(a)) \cdot g'(a)$$

(product of complex numbers) where $f'(g(a))$ denotes the complex derivative of $f$ at the complex number $g(a)$.

By the way, your proof of the single variable chain rule doesn't make much sense. Since $g$ is continuous, you know that $g(x) \to g(a)$ when $x \to a$, but as you noticed you may have $g(x) = g(a)$ for $x \ne a$ so that the quotient is undefined. See any book on calculus how to give a correct proof.