Cauchy Integral Formula - Multiple Possible Solutions?

[Leveret]

Cauchy Integral Formula -- Multiple Possible Solutions?

I'm in the process of teaching myself some complex analysis, and I ran into the following conundrum:

Homework Statement

Suppose one was given an integral,
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z$$
where C is a closed curve with a and b in its interior, oriented counterclockwise.

Homework Equations

$$f(a) = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-a} \mathrm{d}z$$

The Attempt at a Solution

It seems to me that one could approach this one of two ways, and thereby obtain one of two different solutions. One possibility would be to say:

$$f(z) = \frac{2 \pi i}{z-b}$$
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-a} \mathrm{d}z = f(a) = \frac{2 \pi i}{a-b}$$

However, it seems equally valid to say:

$$f(z) = \frac{2 \pi i}{z-a}$$
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-b} \mathrm{d}z = f(b) = \frac{2 \pi i}{b-a}$$

which differs from the former solution by a factor of -1. Is this by design, or am I missing something important about the Cauchy integral formula (which is completely plausible)?

 

[Ray Vickson]

I'm in the process of teaching myself some complex analysis, and I ran into the following conundrum:

Homework Statement

Suppose one was given an integral,
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z$$
where C is a closed curve with a and b in its interior, oriented counterclockwise.

Homework Equations

$$f(a) = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-a} \mathrm{d}z$$

The Attempt at a Solution

It seems to me that one could approach this one of two ways, and thereby obtain one of two different solutions. One possibility would be to say:

$$f(z) = \frac{2 \pi i}{z-b}$$
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-a} \mathrm{d}z = f(a) = \frac{2 \pi i}{a-b}$$

However, it seems equally valid to say:

$$f(z) = \frac{2 \pi i}{z-a}$$
$$\int_C \frac{1}{(z-a)(z-b)} \mathrm{d}z = \frac{1}{2 \pi i} \int_C \frac{f(z)}{z-b} \mathrm{d}z = f(b) = \frac{2 \pi i}{b-a}$$

which differs from the former solution by a factor of -1. Is this by design, or am I missing something important about the Cauchy integral formula (which is completely plausible)?

The Cauchy formula says that if f is homomorphic in the interior of C then your first result would hold: that is, the integral of f(z)/(z-a) would give 2*pi*f(a). However, f(z) = 1/(z-b) is not holomorphic (it has a pole at z = b in the interior of C) so the result need not hold.

RGV

 

[kru_]

Cauchy's Integral Formula requires f(z) to be analytic inside and on the boundary of the contour.

When there are multiple singularities within the contour, we use the Residue Theorem.

 

[Leveret]

Gotcha, thanks!