We sometimes talk about the zeros of a polynomial. Well - a polynomial is analytic, and it has finitely many zeros, where p(z) = 0. The nomenclature is consistent.
A pole is a point p at which f need not be differentiable, or even defined, yet f(z) is analytic about p. A pole is removable if f can be defined at p so that f becomes analytic at p. A pole at p is an mth order pole if zmf(z) makes the pole removable, and m is minimal. Equivalently, take the laurent series about p, and (z-p)-m is the first term. Let's show this in both directions.
Set p = 0 for convenience. If the laurent series starts with sz-m, multiply through by zm and find a power series that is convergent, and analytic throughout the disk, with f(0) = s. Thus setting f(0) = s "removes" the pole. A smaller value of m leaves reciprocals in the laurent series that approach infinity as z approaches 0. We can't make f continuous at 0, much less analytic, hence m is minimal.
Conversely, let f have a pole of order m at 0. Let g = zmf, and set g(0) = s, so that g is analytic. The power series for g begins with the constant term s, and when we divide by zm to get f, the first term is sz-m. But what if s = 0? Then g/z is also analytic, and the pole actually has an order less than m, which is a contradiction. Thus the first term of the laurent series has exponent -m.
A simple pole has order 1.
An essential pole can never be removed. Let f = E1/z. In the previous section we showed Ez is analytic, hence f is the composition of analytic functions, and is analytic everywhere except 0.
Let z approach 0 from the positive x axis. As x gets small, f(x) approaches infinity at an exponential rate. Even p(x)f(x), for some polynomial p, still approaches infinity. If we consider zmf(z), we cannot make this function continuous at 0, much less analytic. Therefore the pole at 0 is an essential pole.
As a corollary, there is no laurent series for f, even though it is analytic everywhere except 0. The laurent operator remains valid, producing a series that converges to f, but the series must have infinitely many terms with negative exponents. This is a reverse laurent series. You can probably guess the series for f. We have the series for Ez, and it converges for all z, so if we apply the series to 1/z it must converge to E1/z.
f(z) = 1 + 1/z + 1/(2z2) + 1/(6z3) + 1/(24z4) + …
While we're on the subject, we can prove that the representation of f as a reverse laurent series is unique, and equal to the series produced by the laurent operator. The trick is reciprocal substitution, as shown above. If f is analytic through an annulus, and f is equal to the sum of a reverse laurent series, set g = f(1/z), which is also analytic through an annulus. The forward laurent series, produced by substituting 1/z for z, converges to g(z), hence g has a laurent series, and it is unique. Therefore the reverse laurent series for f is unique.
Since the laurent operator generates the laurent series for g, reverse this series, and the laurent operator generates a reverse laurent series for f. There is but one reverse laurent series for f, so the series generated by the contour integrals must be the reverse laurent series for f.
With this in hand, the series we wrote for E1/z is the series, and the appropriate contour integrals around the inner circle would indeed produce these coefficients.
There may be, however, other bidirectional series that sum to f. Let f1 be 1/(1-z), with the familiar forward laurent series 1 + z + z2 + z3 + …, and let f2 be f1(1/z), with series 1 + 1/z + 1/z2 + 1/z3 etc. Add these two series to find a bidirectional series that sums to f1 plus f2. After some algebra, f1+f2 = 1, so the constant series 1 would do just as well, at least on the annulus between 0 and 1. The representation of f as a unidirectional laurent series is unique, but other bidirectional series may also sum to f.