I have this proof for finite points but how would I modify it for infinite many points between [2,5]?
Assume $q(z)$ is any function that is holomorphic on a disc U except at a finite number of points $\xi_1,\ldots, \xi_n\in U$, and assume $\lim_{z\to\xi_j}(z-\xi_j)q(z)=0$ for $1\leq j\leq n$. Let $U'=U-\{\xi_1,\ldots\xi_n\}$. Then q is holomorphic on U'.
Note $F(z)=q(z)-f'(a)$ so $q(z)=\frac{f(z)-f(a)}{z-a}$
Step 1 is the Cauchy-Goursat argument:
$\int_{\partial R}q(z)dz=0$ for all rectangles R in U such that $\xi_j\notin\partial R$ for j.
proof:
Let R be a rectangle with the boundary of R in U'. First subdivide R into sub-rectangles so that each sub-rectangle has at most one $\xi_j$ inside it. By Cauchy-Goursat, $\int_{\partial R}=\sum_i\int_{\partial R_i}$. So it suffices to show $\int_{\partial R}q=0$ if R contains at most one $\xi_j$.
If R contains no $\xi_j$, then we are done by Cauchy-Goursat. Assume $\xi=\xi_j$ is inside R. Let $\epsilon>0$ be given. Put $\xi$ in a square of size x at the center of this where x is chosen small enough so that $|(z-\xi)q(z)|<\epsilon$ for z in and on this square. Subdivide this rectangle so the square which is x by x is its own rectangle around $\xi$. As before, the integrals of the sub-rectangles that don't contain $\xi$ are 0. So $\int_{\partial R}q=\int_{\text{square with xi}}q$.
$$
\left|\int_{\text{square with xi}}q\right|\leq \underbrace{||q||_{\text{square with xi}}}_{\frac{\epsilon}{\min\{|z-\xi|\}}}\times \underbrace{(\text{length of square with xi}}_{4x}\leq\frac{\epsilon}{x/2}4x=8\epsilon
$$
So $\left|\int_{\text{square with xi}}q\right|=0$.
Step 2 is to use step one to create a primitive for q on all of U.
Define $g(a)=\int_{z_0}^{z_1}q(z)dz$ where the path is from $z_0$ horizontal and then vertical. So $g(a)$ is well-defined. If a point is not unreachable, then $\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=q(a)$ by exactly the same means as before. Unreachable are the points vertical above $\xi$.
When computing $\frac{g(a+h)-g(a)}{h}$, only consider h in C with $|h|<\frac{\delta}{2}$. The path for computing g(a+h) also misses $\xi$. Then for these h $g(a+h)-g(a)=\int_a^{a+h}q$ so the same reason as before show $\frac{g(a+h)-g(a)}{h}-g(a)\to 0$ as $h\to 0$.
To handle all the bad exception.
Pick $\epsilon\geq 0$ such that the point $z_1=z_0+\epsilon(1+i)$ is not on any of the same vertical or horizontal lines as any $\xi_j$. Of course epsilon is really small compared to the radius of U.
Define $g_1(a)=\int_{z_1}^aq(z)dz$. This defines a primitive for q(z) on all of the disc except on the unreachable regions.
Define $g_2(a)=\int_{z_1}^aq(z)dz$ but this time move vertical and then horizontal. So $g_2(a)$ is also a primitive for q on U except at the its unreachable points (but g_1 reaches those points and vice versa). On the overlap, $g_1$ and $g_2$ are primitives for the same q. They differ only by a constant. But $g_1(z_1)=0=g_2(z_1)$ so $g_1=g_2$ on the overlap. For a small enough epsilon,
$$
g(z)=\begin{cases}g_1(z)\\g_2(z)\end{cases}
$$ is defined on all of U' and is a primitive for q.
How do I extend this to my problem?
