In our lecture notes for our complex analysis class, we were given the following theorem:
Theorem: There exists a unique complex function $f$ such that
- $f(z)$ is a single valued function $f(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ and $f(1)=e$.
- $\forall z_{1},z_{2} \in \mathbb{C}$, $f$ satisfies $f(z_{1}+z_{2}) = f(z_{1})f(z_{2})$.
- $f$ is complex differentiable for all $z \in \mathbb{C}$. Such $f(z)$ is denoted by $e^{z}$ or $\exp(z)$ and called the exponential function.
On our homework, we were asked to give an alternative form of the proof by following these steps:
(a) Prove that $f^{\prime}(0)=1$. (Hint: Argue why the limit $\lim_{\Delta z \to 0} \frac{f(\Delta z)-f(0)}{\Delta z}$ must be the same as $\lim_{\Delta x \to 0} \frac{f(\Delta x)-f(0)}{\Delta x}$, $x \in \mathbb{R}$.)
(b) Prove that $f^{\prime}(z) = f(z)$ for all $z \in \mathbb{C}$.
(c) Treating the resulting differential equation $f^{\prime}=f$, ,$f(0)=f^{\prime}(0)=1$ as a system of real-number differential equations, prove that the only complex differentiable function $f$ that satisfies it is $f(z) = e^{x}(\cos y + i \sin y)$. (Hint: Don't forget the Cauchy-Riemann equations.)
I think that once I have part (a) taken care of, I can figure out (b) and (c) by myself, but I am having a lot of trouble with part (a) - even understanding how arguing that $\lim_{\Delta z \to 0} \frac{f(\Delta z)-f(0)}{\Delta z}$ must be the same as $\lim_{\Delta x \to 0} \frac{f(\Delta x)-f(0)}{\Delta x}$, $x \in \mathbb{R}$ would help us, and how even to do so.
Basically, as far as I've gotten is the following:
Suppose $f$ is such a function.
Then $f^{\prime}(0) = \lim_{\Delta z \to 0} = \frac{f(\Delta z) - f(0)}{\Delta z}$.
And since $\Delta z = z - z_{0} = z-0 = z$, this becomes $\lim_{z \to 0} \frac{f(z)-f(0)}{z} = \displaystyle \lim_{(a+bi) \to (0+0i)}\frac{f(a+bi)-f(0)}{a+bi} = \displaystyle \lim_{(a+bi) \to (0+0i)}\frac{f(a)f(bi) - f(0)}{a+bi}$.
But, I have no idea how to take it from there, or even if what I've done so far is correct.
Could someone please help me with part (a)? Once I have that done, I'm pretty sure parts (b) and (c) shouldn't be a problem.
Thanks.
It follows directly from the definition of limit. You want to show $$\lim_{z \rightarrow 0} \frac{f(z) - f(0)}{z} = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$$ where $x = \mathfrak{Re}(z)$. Now $z$ can approach $0$ from any direction, and provided the complex limit exists, value of the limits corresponding to all these direction must be same. In particular, restrict the direction to the real axis.
That the limit exists follows from condition $(3)$.