Precourse worksheet I Complex analysis

Recall the Cauchy--Riemann equations for differentiable
functions f(z) of a complex variable z = x+iy. A function f(z)
that is differentiable on an open domain is _holomorphic_
(or regular, or complex analytic).

Suppose f(z) is differentiable as a function of z in a neighbourhood
of a point, and write
   df = (partial df/dx)dx + i(partial df/dy)dy.
Show that the contour integral of df around the infinitesimal square
   [x,y], [x+delta*x, y], [x+delta*x, y+delta*y], [x, y+delta*]
is zero to first order. Recall the Cauchy integral theorem. Look
up a proof if you can't remember it.

Calculate the contour integral of dz/z around the anticlockwise unit
circle. [Hint: Use polar coordinates (r, theta). Note that
dz/z = d(log z).]

Let f(z) be a holomorphic function on the unit disc. Use the Cauchy
integral theorem to prove that for a in the disc, the value f(a) is
equal to the contour integral of f(z)dz/(z-a) around the unit circle.

A holomorphic function f(z) is locally a power series
   f(z) = a_0 + a_1z + ..
with a nonzero radius of convergence. This was probably surprising
when you first saw it after all the fuss about the Taylor series of
a real function that is differentiable to order 1, 2, .., infinity.
Prove that the real function exp(-x) is C^\infty (infinitely
differentiable) at x=0, but is not equal to its Taylor series.

A holomorphic function f(z) has zero of order d at (z=0) if the
coefficients of its power series start a_0 = a_1 .. = a_{d-1} = 0,
followed by a_d ≠ 0

A _meromorphic function_ is a ratio f = g/h of two holomorphic
functions, with h nonzero. If g has zero of order a and h has zero of
order b then f = g/h has pole of order (b-a). A meromorphic function
with pole of order d has a Laurent series
   f(z) = a_{-d}/z^d + .. a_{-1}/z + a_0 + a_1z + ..
The negative terms make up the _principal part_ of f(z). Allowing
a pole of order <= d gives you a vector space of dimension d of
principal parts.

The _residue_ of the above f at 0 is the coefficient a_{-1} of
1/z in its Laurent series. It is given by the Cauchy contour integral
of f*dz around a small (anti-clockwise) circle at 0.

In terms of commutative algebra, the ring of functions that are
holomorphic at 0 is a discrete valuation ring DVR, and the ring of
meromorphic functions is its field of fractions.

Prove the maximal modulus principle: if f is a holomorphic function
on a disc |z| <= r (which is compact, of course), the real valued
function |f(z)| takes its maximal value at a boundary point
(with |z| = r).

Deduce Liouville's theorem I: a bounded holomorphic function on the
entire complex plane CC is constant.

Now let f(z) be a function defined and holomorphic on the
punctured disc 0 < |z| <= 1. Suppose also that f(z) has at most
polynomial growth of order d as z -> 0; in other words,
|z|^d*|f(z)| is bounded on the unpunctured disc |z| <= 1. Prove
that the function g(z) = z^(d+1)*f(z) extended by g(0) = 0 is
continuous and holomorphic on the disc. Corollary: polynomial
growth at P implies meromorphic at P.

Write PP^1(CC) for the complex plane extended by a single point
\infty at infinity (also referred to as the closed complex plane
or the Riemann sphere). Prove Liouville's theorem II: a holomorphic
function on the entire complex plane CC that has polynomial growth
of order <= d as z -> infty is a polynomial of degree <= d. [Hint:
write w = 1/z for a local parameter near \infty and use the result
of the previous section.]

Prove that polynomials of degree <= d form a CC-vector space of
dimension of dimension d+1.

If a,b in CC, prove that (z-a)/(z-b) is a meromorphic function
on PP^1(CC) with zero of order 1 at a and pole of order 1 at b.