I want to prove that if $\mathcal{U}\subset \mathbb{C}$ is an open set, $f:\mathcal{U}\longrightarrow \mathbb{C}$ is differentiable at $z_0\in\mathcal{U} \Longleftrightarrow \exists A \in \mathbb{C}$ and a function $\epsilon(z)$ so that:
$$f(z)=f(z_0)+A(z-z_0)+\epsilon(z)(z-z_0)$$
Actually, I am only interested in the backwards implication (I have already proved the other one). My approach is this: I am using the deffinition of differentiable at a point and substituting the value of $f(z)$ by the one above, trying to see if the limit exists. Therefore I am trying to compute:
$$\lim_{h\rightarrow0}\frac{f(z+h)-f(z)}{h}=
\lim_{h\rightarrow0}\frac{f(z_0)+A(z-z_0)+\epsilon(z)(z-z_0)-f(z_0)-A(z+h-z_0)-\epsilon(z+h)(z+h-z_0)}{h}$$
After doing some calculations I get:
$$\lim_{h\rightarrow0}A+\epsilon(z+h)+(z+z_0)\frac{\epsilon(z+h)-\epsilon(z)}{h}$$
My problem is that I am not sure if this last limit exists, since I don't know if $\epsilon$ if differentiable at every $z\in\mathcal{U}$. How can I finish the proof? I know that maybe I could use Taylor for a different approach of my problem, but this is my first course in complex analysis and my teacher hasn't yet defined Taylor series for complex functions, so please avoid that in your answers.
Prove that $f(z)=f(z_0)+A(z-z_0)+\epsilon(z)(z-z_0) \Rightarrow f$ differentiable
complex-analysisderivativeslimits
Best Answer
That statement is false, unless you add something about $\varepsilon$, such that $\lim_{z\to z_0}\varepsilon(z)=0$. Then, asserting that $f$ is differentiable at $z_0$ means that the limit $\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}h$ exists. But\begin{align}\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}h&=\lim_{t\to0}\frac{Ah+\varepsilon(z_0+h)h}h\\&=A+\lim_{h\to0}\varepsilon(z_0+h)\\&=A.\end{align}