# Complex analysis

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Consider a domain D and f:D-->$\mathbb{C}$ a holomorphic function and C a contractible Jordan path contained in D and z1, z2, two points in the interior of C. Evaluate

$$\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz$$

What happens as $z_1 \rightarrow z_2$?

I have found that

$$\int_C \frac{f(z)}{(z-z_1)(z-z_2)}dz = \frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))$$

and that is correct according to the solution manual.

Of course, as $z_1 \rightarrow z_2$, the RHS goes to $2i\pi f'(z_2)$, and according to the solution manual, the LHS goes to

$$\int_C \frac{f(z)}{(z-z_2)^2}dz$$.

But considering the integral as one big function of z_1 (let's call it $I(z_1)$), since it equals

$$\frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))$$,

which is not defined, hence not continuous at $z_1 = z_2$, I cannot say that

$$\lim_{z_1\rightarrow z_2} I(z_1) = I(\lim_{z_1\rightarrow z_2} z_1)$$

So how is the conclusion attained?

thx!

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## Answers and Replies

HallsofIvy
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The conclusion is exactly what you said to start with:
That, as z1 goes to z2, the integral goes to $2\pi i f '(z_2)$ .

shmoe
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quasar987 said:
But considering the integral as one big function of z_1 (let's call it $I(z_1)$), since it equals

$$\frac{2i \pi}{z_1-z_2}(f(z_1)-f(z_2))$$,

which is not defined, hence not continuous at $z_1 = z_2$, I cannot say that

It doesn't matter that it isn't defined when $$z_1=z_2$$, it's a limit and we don't care what actually happens at this point.

Hurkyl
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So what have you found?

You have an expression with a limit at z1=z2. However, the expression is undefined at z1=z2. This is a situation you've seen in calc I!

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Like schmoe said, in regular situations, it doesn't matter if the function is defined for the limit to exists. But you can't use $\lim_{x\rightarrow a}f(x) = f(\lim_{x\rightarrow a}x)$. So I don't know why I can bypass in integral, and write

$$\lim \int = \int \lim$$

like the solution manual suggest i do.

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It could also be that there is no a priori way of knowing if I can "bypass" the integral, but the fact that the derivative of a Cauchy integral is itself a Cauchy integral let us conclude:

$$\frac{2i\pi}{1!} f'(z_2) = \int_C \frac{f(z)}{(z-z_2)^2} dz$$

But all of you people seem to be saying there is a way of knowing if I can bypass the integral, and it is suposeldly very simple if it is Calc I material. Hint plz?

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Hurkyl??

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Hurkyl
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From Advanced Calculus by Buck:

Theorem If the functions $f_n$ and F are integrable on a bounded closed set E, and $\{ f_n \} \rightarrow F$ pointwise on E, and if $||f_n||_E \leq M$ for some M and all n = 1, 2, ..., then $\lim_{n \rightarrow \infty} \int_E f_n = \int_E F$.

There are various theorems that look like this. You might even be able to use this one for your purposes. (The norm $||f||_E$ means the maximum of the absolute values of the function on the set E)

Hopefully you've had them before! Does your complex analysis book talk at all about interchanging limits and integrals? Or maybe you've had a real analysis / calculus course that talked about it? (One buzzword is "uniform convergence")

You could even try proving it yourself that you can interchange the integral and the limit in this case. But the observation in post #6 should serve as a substantial shortcut.

(P.S. I read your OP as saying that you were having problems with the fact the RHS was undefined when you let z_1 -> z_2, and I was referring to that)

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shmoe
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Hurkyl said:
(P.S. I read your OP as saying that you were having problems with the fact the RHS was undefined when you let z_1 -> z_2, and I was referring to that)

This is how I read it as well.

quasar987, you've proven the integral you started with converges to the derivative and you also have the alternate expression in post 6 for this derivative. There's no need here to swap the integral with the limit. A substantial shortcut as Hurkyl suggested.