Questions about poles and residues (complex analysis)

I mean, clearly if we have a fraction with (z-a) on the bottom then a will be a pole. But what if we have a fraction with a 1/z term or, say (1 + 1/z)^2 on the top? Is that going to give a pole at 0?

How can I tell the order of a pole when it's not completely clear? Say we have a pole at a. Do I just take my function f(z), and take the function gn(z) = (z-a)^n * f(z) for each natural number n, and find the least natural number n such that gn(a) is not zero?

What is the easiest/most straightforward way to find residues? Should I try to get at the -1 term of the Laurent expansion using Taylor/power series expansions, or will it usually be easier to just use the derivative formula
http://en.wikipedia.org/wiki/Residue_(complex_analysis)#Limit_formula_for_higher_order_poles
?

More or less. Formally, if you have a holomorphic function $f : U \to \mathbb{C}$ with U containing an open annulus around z, we say f has a pole of order n at z if $\lim_{\zeta \to z} (\zeta - z)^n f(\zeta)$ exists and is non-zero.



$\displaystyle \left( 1 + \frac{1}{z} \right)^2 = \frac{(z + 1)^2}{z^2}$, so this has a double pole at 0.



Yes.



Do a few practice questions and see which is easier for you. If you know in advance that you have a simple pole you can take a shortcut and try to directly evaluate $\lim_{\zeta \to z} (\zeta - z) f(\zeta)$, for example.

Thank you.

If I want to evaluate a real integral with infinity in one (or both) of its limits, I know that I would use the usual semi-circular contour. In these cases, is the integral round the semi-circular arc always going to go to 0? In all of the examples I've seen, this seems to be the case.

By no means! Sometimes you will see an integral with something like sinh x in the denominator, in which case, not only would you not have convergence on a large semicircle, but you would be adding an infinite series of residues as the radius of the semicircle increased!

Look out for a complex analysis book with plenty of examples and you should see some instances of how to use a rectangular contour for example.

It seems that if Jordan's lemma applies, then I should go for the semi-circle and that otherwise I should use the rectangle (and the integrand will usually be something involving exponentials).

How do I deal with the indented semi-circle contour, where we go round the origin in a semi-circle to avoid a point of non-holomorphy? But what happens if I get something/r in the integrand (say exponential/r)? Does this mean that I have definitely done something wrong? This kind of integrand would appear to blow up if we let r->0.

I'm having trouble understanding the keyhole contour.

http://www2.imperial.ac.uk/~bin06/M2PM3-Complex-Analysis/m2pm3l30(11).pdf

Check near the end of the document. It says "Near gamma 3: argz = 2pi.....". Why is it important that we parametrized it as z = x* e^(2*pi*i)? It clearly made a difference somewhere but why didn't we just replace that e^(2*pi*i) by 1? How can we have an upper branch where the argument is 0 and a lower branch where the argument is 2*pi? I mean, aren't they the same? We could perhaps think of them as slightly different but won't we still get the same things coming out when we plug in 2*pi instead of 0? So I don't understand why this method works, when we would expect the upper integral and lower integral (above and below the axis) to cancel each other out and thus not give us anything useful.

They don't cancel as on the lower side of the branch cut you have $e^{2ai\pi}$ time the 1 where $0 < a < 1$ so you will get

$(1- e^{2ai\pi})I =$ sum of residues

The LHS is not zero as a is not an integer.

But if we had parametrized it as z = x instead of z = x*exp(2*pi*i), we would get -I wouldn't we? But why? x = x*exp(2*pi*i) !

Because you have cut the plane along the positive real axis, $0\leq argz < 2\pi$, you take $\gamma_3 = \displaystyle\lim_{\delta\to 0}x - \delta i$.

As you have cut the plane along the real axis the arguement can't jump back to zero

I thought that, wherever one sees exp(2*pi*i), this can just be replaced by 1.

No, you can't always do this.

Why not? Aren't they equal?

I've another question. What is the residue of f(z) = 1/z^2 * pi * cot(pi*z) at 0? I tried using the derivative rule by letting g(z) = z * pi *cot(pi*z) [since 0 is a triple pole] and differentiating twice. But then when you plug in 0 we get division by zero.

Not when you've got a fractional power in the mix, no.

You can't "plug in 0", you need to take the limit as z->0.