Q:

5.16 Remarks from baby Rudin

How Rudin got expression $(29)$? I think so: If $f(t)=f_1(t)+if_2(t)$ then $$f'(x)=\lim_{t\to x}\dfrac{f(t)-f(x)}{t-x}=\lim_{t\to x}\dfrac{f_1(t)-f_1(x)}{t-x}+i\lim_{t\to x}\dfrac{f_2(t)-f_2(x)}{t-x}=f_1'(x)+if_2'(x).$$I think that it's not correct because we don't know does these limits exists?
Can anyone explain this moment please?

A:

Let $f:[a,b]\to \mathbb{C}$ is a complex function. Then $f(t)=f_1(t)+if_2(t)$ for $t\in [a,b]$ where $f_1, f_2$ are the real.
Theorem: $f$ is differentiable at $x$ iff both $f_1$ and $f_2$ are differentiable at $x$.
Proof:
$\Leftarrow$ If both $f_1$ and $f_2$ are differentiable at $x$. Then exists limits $\lim \limits_{t\to x}\dfrac{f_i(t)-f_i(x)}{t-x}=f'_i(x)$ for $i=1,2$. Summing these we get: $$\lim \limits_{t\to x}\dfrac{f(t)-f(x)}{t-x}=f'_1(x)+if'_2(x).$$ So $f'(x)=f'_1(x)+if'_2(x).$
$\Rightarrow$ Let $f$ is differentiable at $x$. Then exists limit $$f'(x)=\lim \limits_{t\to x}\dfrac{f(t)-f(x)}{t-x}=\lim \limits_{t\to x}\left [\dfrac{f_1(t)-f_1(x)}{t-x}+i\dfrac{f_2(t)-f_2(x)}{t-x}\right].$$ Let $\operatorname{Re}f'(x)=u(x)$ and $\operatorname{Im}f'(x)=v(x)$. From inequality: $$0\leqslant \left|\dfrac{f_1(t)-f_1(x)}{t-x}-u(x)\right|\leqslant \left|\dfrac{f(t)-f(x)}{t-x}-f'(x) \right|$$ we get that $u(x)=f'_1(x)$ and also $v(x)=f'_2(x)$. Q.E.D.

