Let f and g be represented by laurent series over an annulus. If h = f×g, we will show that the polynomial product, the series for f times the series for g, yields the series for h. Since there is at most one series for h, it is enough to show that the polynomial product converges to h.
If the series for f times the series for g yields the series for h, then the same is true for zf times g. We already know that multiplication by z increments the exponents on the laurent series. So the exponents on the series of f increase, and this in turn causes the exponents on the product series to increase, and sure enough, the product series converges to zh. Similarly, f can be divided by z, and the polynomial product will produce the series for h/z. This allows us to advance f and g from laurent series to power series. Prove this theorem for power series, and it applies to laurent series as well.
Let f and g be analytic about 0, with their respective power series, and let t be the radius of the smaller circle of convergence. Let z be a point with norm less than t. Remember that f and g converge absolutely at z. Replace the terms of the series of f(z) with their norms, giving a series u, at the point z. Remember, u is absolutely convergent. similarly, let v hold the norms of the terms of g(z).
Build a matrix M such that Mi,j = uivj. This is a two dimensional series, and the sum of M is the product of the sum of u timese the sum of v. Thus M converges absolutely.
Replace uivj with figj, giving a complex number that is dominated by uivj. Thus the cross product of the series of f times the series of g converges absolutely.
We can add terms in any order, so use the standard method for multiplying two polynomials. Here is a simple example.
f = 1/(1-z)
f = 1 + z + z2 + z3 + z4 + …
g = 1/(1-z2)
g = 1 + z2 + z4 + z6 + z8 + …
h0 = 1×1 = 1
h1 = 1×z = z
h2 = 1×z2 + z2×1 = 2z2
h3 = 1×z3 + z2×z = 2z3
h4 = 1×z4 + z2×z2 + z4×1 = 3z4
h50 = 26z50
h = 1/(1-z-z2+z3)
So far we showed the product series converges to something, but does it converge to h? Since it converges inside t it is analytic inside t. If we can show the derivatives are the same as the derivatives of h, then the two functions are the same, and we're done.
Start with the 0 derivative, which is the value of h at 0. This is f(0)g(0), which agrees with the product of the constant terms in the two series.
The first derivative is f′g+fg′. Represent f and g as series, and their first derivatives are f1 and g1, the linear coefficients. Plug these into the product rule and get f1g0+f0g1. This agrees with the linear coefficient on z in the polynomial product.
Apply the product rule twice, and the second derivative is f′′g+2f′g′+fg′′. When f is a series, its second derivative is 2f2. The first derivative is still f1 as before. Substitute these values in the product rule to get 2f2g0+2f1g1+2f0g2. This is twice the coefficient on z2 in the polynomial product, which happens to be the second derivative of the product series.
Let me be bold and jump ahead to the jth derivative. Apply the product rule j times and find an expression that looks like the jth row of pascal's triangle. In this row, the fourth derivative of f, times the j-4 derivative of g, is multiplied by the binomial coefficient j choose 4, which is j! over 4!×(j-4)!. Evaluate the fourth derivative of the series f and obtain 4!×f4. Similarly, the j-4 derivative of g brings in a factor of (j-4)!. These conspire to clear the denominator, and the coefficient on the fourth term becomes j!. The same thing happens to the other terms in our expression. The result is j! times the jth coefficient in the polynomial product, which happens to be the jth derivative. The derivatives agree out to infinity, and the polynomial product is indeed the series for h.
If f g and q have convergent laurent series, and the series for q times the series for g produces the series for f, we can multiply q and f by z, and shift their series by 1, and the relationship still holds. Similarly, we could divide g and f by z, and the relationship still holds. Use this fact to adjust f and g, so that they are both power series with nonzero constant terms. Now f and g are analytic on the entire disk, and f(0) and g(0) are nonzero.
Pull back to a neighborhood about 0, such that g is nonzero throughout. Thus q = f/g is well defined and analytic on this (smaller) disk. Write a power series for q and use the above to show the series for q times the series for g equals the series for f. The series for q exists, and is unique, and can be computed via synthetic division. This is a form of long division where one laurent series is divided into another.
Note that the quotient series need not converge on our original annulus. Granted, we have found the quotient series for q, but it's circle of convergence may not reach this far. For instance, let f = 1 and g = 1-z. Both functions are analytic and nonzero on an annulus whose inner radius is 2 and outer radius is 3. The quotient is analytic, with a power series of 1 + z + z2 + z3 + z4 + …, however, this series does not converge past a radius of 1. If you use contour integrals to find the true series for 1/(1-z), converging outside of the unit circle, it will look quite different.
Rarely do we get the chance to compute contour integrals by hand, so let's do that now. You'll see that it matters, whether we evaluate these integrals inside or outside the unit circle.
Remember that all the integrals are based on f(y)yk, or in this case, yk/(1-y). Start with k positive and write it this way.
1/(1-y) = 1 + y + y2 + y3 + … + yk/(1-y)
∫ yk/(1-y) = -log(1-y) - y - y2/2 - y3/3 - …
Evaluate this around a circle centered at the origin. We start and stop at the same y, so all the terms drop out except for -log(1-y). If the circle is entirely to the left of 1, the angle of 1-y returns to its original position, and the integral is 0. If the circle encloses the pole at 1, i.e. if the radius of our circle exceeds 1, the log has increased its angle by 2π. Fold in the constants, and the coefficients on the negative powers of z are -1, when the radius exceeds 1, and 0 otherwise.
The constant term comes from the integral of f(y)/y, or 1 over y times (1-y), or 1/(1-y) + 1/y. The integral becomes log(y)-log(1-y), and log(y) always increases its angle by 2π, no matter the radius, so the constant term is 1 inside the unit circle and 0 outside.
The next integrand has another y in the denominator, like this.
1 / (y2-y3)
1 / (y-y2) + 1/y2
We just calculated the integral of the first term. The integral of the second is proportional to 1/y, which returns to its starting point, and is 0. The linear coefficient is 1 inside the unit circle and 0 outside. This continues by induction; the rest of the coefficients are all 1 or 0. In side the unit circle we find the expected power series 1 + z + z2 + z3 etc. Outside the unit circle we have the series -1/z - 1/z2 - 1/z3 - 1/z4 etc. This is geometric, and converges to a value we will call s. If w and z are reciprocal, 1-s = 1/(1-w). Replace w with 1/z and 1-s = z/(z-1). Solve for s and get 1/(1-z) as expected.