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## Holomorphic on C

 Quote by fauboca Does it have to do with the winding number?
Not really. It has to do with allowing more general $\gamma$ instead of just circles.

Like I said in an earlier post, this approach is tricky. Morera's theorem is definitely the way to go here.
 We have a theorem we call our q-theorem. Suppose q(z) is any function that is holomorphic on a disc U except at a finite number of points in U, and assume the $\lim_{z\to\xi_j}(z-\xi_j)q(z) = 0$. Where $\xi_j$ are those finite points. Then q has a primitive and is analytic inside the disc. $$q(z) =\frac{f(z)-f(z)}{z-a}$$ But this only for a finite number of points and I have an infinite number.
 Let $x\in[2,5]$. Then $$g(b) = \frac{1}{2\pi i}\int_C\frac{f(u)}{u-b}du,$$ i.e. $g = f$ for all $b\neq x$ where x is a removable singularity. Now g is holomorphic everywhere inside C so g is the extension of f to the interior of C. Then let g=f on $\mathbb{C}$ too.

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 Quote by fauboca Now g is holomorphic everywhere inside C so g is the extension of f to the interior of C.
No - you still need to show that g(z)=f(z) for every z in C\[2,5].

 Quote by morphism No - you still need to show that g(z)=f(z) for every z in C\[2,5].
I don't know what to do.

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 Quote by fauboca I don't know what to do.
You're better off just abandoning this approach and using Morera's theorem for a more elegant solution.

But for the sake of completeness, this is how we could've wrapped up. Let $\gamma_\epsilon$ be a small $\epsilon$-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient $\gamma_\epsilon$ clockwise and C counterclockwise so that
$$f(z) = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + \int_C \frac{f(w)}{w-z} dw = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + g(z)$$ for all z in between $\gamma_\epsilon$ and C. This holds for all small $\epsilon$, so by letting $\epsilon \to 0$ and using the continuity of f, we see that the the integral over $\gamma_\epsilon$ in the above equations goes to zero, and consequently f(z)=g(z) for all z in C\[2,5].

 Quote by morphism You're better off just abandoning this approach and using Morera's theorem for a more elegant solution. But for the sake of completeness, this is how we could've wrapped up. Let $\gamma_\epsilon$ be a small $\epsilon$-band around [2,5] fully contained in C (here I'm assuming the radius of C is > 1.5). Orient $\gamma_\epsilon$ clockwise and C counterclockwise so that $$f(z) = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + \int_C \frac{f(w)}{w-z} dw = \int_{\gamma_\epsilon} \frac{f(w)}{w-z} dw + g(z)$$ for all z in between $\gamma_\epsilon$ and C. This holds for all small $\epsilon$, so by letting $\epsilon \to 0$ and using the continuity of f, we see that the the integral over $\gamma_\epsilon$ in the above equations goes to zero, and consequently f(z)=g(z) for all z in C\[2,5].
How is it done via Morera?

Doesn't Morera only say we have a primitive and closed curves are 0. How does that make it an easier approach?
 Recognitions: Homework Help Science Advisor Morera tells you that f must be holomorphic in a region G if $\int_\gamma f = 0$ for all closed curves $\gamma$ in G. The converse is of course Cauchy's theorem. So you just need to show that $\int_\gamma f = 0$ for certain curves $\gamma$.

 Quote by morphism Morera tells you that f must be holomorphic in a region G if $\int_\gamma f = 0$ for all closed curves $\gamma$ in G. The converse is of course Cauchy's theorem. So you just need to show that $\int_\gamma f = 0$ for certain curves $\gamma$.
How is that shown?
 Recognitions: Homework Help Science Advisor Try to think it over for a bit...

 Quote by morphism Try to think it over for a bit...
By Goursat's rectangle method?

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 Quote by fauboca By Goursat's rectangle method?
Yes, something like that would work.

 Quote by morphism Yes, something like that would work.
I am still lost on how to do this though.

 Quote by morphism Yes, something like that would work.
So I am trying to use Morera's Theorem:
Let U be an open set in C and let f be continuous on U. Assume that the integral of f along the boundary of every closed rectangle in U is 0. Then f is holomorphic.

So let $U = \mathbb{C} - [2,5]$ Let R be rectangles in U which are parallel to the coordinate axes. So $\int_{\partial R}f=0$.

Now how can I use this?
 Recognitions: Gold Member Science Advisor Staff Emeritus By the way, "holomorphic at every point" doesn't make sense- "holomorphic" is not a point property. You may mean "analytic at every point". "Holomorphic" means "analytic at every point in the complex plane".
 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?

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