I’m trying to prove that a function $f: \mathbb C \rightarrow\mathbb C$ continuous at some point $a$ has affine part $Az+B\overline z+C$ if and only if it is differentiable at $a$. Note: The affine part of $f$ satisfies $\lim\limits_{z \to a} \frac{|f(z)-(Az+B\overline z+C)|}{|z-a|}=0$, and I take $\lim\limits_{z \to a} \frac{f(z)-f(a)}{z-a}=w$ to be the definition of $f’(a)$ provided $w$ exists. I have the forward direction of the proof, but I’m stuck trying to show the converse; namely, differentiability at $a$ gives rise to the affine part. Since we assume differentiability at $a$, we have continuity at $a$. Then, my idea is to consider the linear approximation $f’(a)(z-a)+f(a)$ and manipulate it to the form $Az+B\overline z +C$, but I’m stuck doing so. Can anyone get me on the right track?
In my mind, the affine part of the function is the best linear approximation, so it is similar to the line tangent to the graph of a real-valued function at a point. This is why I am attempting to use the linear approximation written above.