The limit of a complex function becoming the limit of 0 (algebraically inside the limit)

Question

For a complex function $f(z)$ holomorphic in $0<|z-a|≤R$ , and given that

(I) $0=\lim_{|z|\to a}(z-a)f(z)$ , (then I did the following)

$=\lim_{|z|\to a}(z-a)[f(z)-f(a)+f(a)]$

$=\lim_{|z|\to a}(z-a)[f(z)-f(a)]+(z-a)[f(a)-f(z)+f(z)]$

$=\lim_{|z|\to a}(z-a)[f(z)-f(a)]+(z-a)[-f(z)+f(a)]+(z-a)f(z)$

$=\lim_{|z|\to a}(z-a)[0]+(z-a)f(z)$

$=\lim_{|z|\to a}(z-a)[0]+0 $ (I)

So, $\lim_{|z|\to a}(z-a)f(z)=\lim_{|z|\to a}(z-a)[0]$

Does that directly imply $f(z)=0$? or $f'(z)=0$? I am trying to prove

$_{|z-a|=r}\int{f(z)}=0$ with $0<r≤R$

Does seeing this at least mean that I am in the right direction before actually proving it? Seeing a counterexample would help to alleviate my curiosity in this if it happens to not mean much.

Answer 1

I suspect that there's a typo somewhere. The question exactly as stated is trivial, at least if $a\ne0$: If $|b|=a$ while $0<|b-a|<R$ then $(b-a)f(b)=0$, hence $f(b)=0$. The set of such $b$ has an accumulation point, so $f=0$.