Section 5.6

Question 1: (a) The function is analytic everywhere except at 0 and -1. The denominator has a double zero at 0 and a simple zero at -1, and the numerator has a simple zero at -1 and no zero at 0, so the quotient has a double pole at 0 and a removable singularity at -1.

(c) 4z is always analytic, so it does not affect any singularities. the denominator has simple zeroes at +1 and -1, while the numerator is non-zero at those points, so the function as a whole has a simple pole at +1 and -1.

(d) e^z - 1 has zeroes at all multiplies of 2pi i, and they are all simple (since the derivative of e^z-1 is non-zero at these points). So the function has simple poles at all multiples of 2 pi i.

(g) The only place where the function is not analytic is at 0. The easiest way to do this one is by finding the Laurent series. Rewriting the function as (sin(3z) - 3z) / z^2, we see that the numerator is - (3z)^3/3! + (3z)^5/5! - ..., so the Laurent series of the function as a whole starts with a linear (z^1) term. In particular, there are no negative power terms, so the function has a removable singularity, and because the first term is z^1, the function has a simple zero once the singularity is removed.

Question 2: 2 cos z - 2 + z^2 has a zero of order 4 at 0, as one can see by a Taylor series expansion (or by differentiation). Thus, the denominator of f has a zero of order 8, and f itself has a pole of order 8 at 0.

Question 3: See examples at back of book.

Question 6: Since f has a pole of order m, its Laurent expansion around z_0 begins with the term a_m/(z-z_0)^m, where a_m is non-zero. Differentiating the Laurent expansion term by term (this is justified for reasons having to do with uniform convergence), we see that f' has a Laurent expansion starting with the term -m a_m/(z-z_0)^{m+1}, which is a non-zero term of order m+1. Thus f' has a pole of order m+1.

Section 5.7

Question 1: (c) This function converges to a non-zero finite number (1, in fact), so it has a removable singularity at infinity. Alternatively, one can replace z by 1/z and look at what happens at 0.

(d) This function goes to zero at infinity, and if you multiply by one power of z, it still goes to zero. If you multiply by two powers of z, it converges to a non-zero finite number. Thus, it has a removable singularity at infinity, and it has a zero of order 2 once the singularity is removed.

(e) This function is the reciprocal of the one in (d), so it must have a pole of order 2. Alternatively, one can note that it goes to infinity at infinity, but once you divide out two powers of z, it converges to a constant. Alternatively, you can replace z by 1/z and look at what happens at 0. Alternatively, you can look at the numerator (triple pole) and denominator (simple pole) and divide one by the other.

(g) This function goes to zero as z tends to infinity when z is real, but don't let that fool you. In the (say) imaginary direction sin z blows up exponentially (it's basically sinh z), and exponential growth beats the 1/z^2 decay. sin z/ z^2 actually has an essential singularity at infinity. Alternatively, note that sin z itself has an essential singularity at infinity (it oscillates along the real axis), and multiplication by a double zero such as 1/z^2 is not going to affect the essential nature of the singularity. Alternatively, you can replace z by 1/z and look at the Laurent expansion around 0.

(h) Since sin z has an essential singularity at infinity, 1/sin z also has an essential singularity at infinity.

Question 6, Solution A. Let C_R be a really big circle of radius R oriented clockwise. By adding and subtracting a contour between Gamma and C_R, one can make a C-shaped contour containing Gamma and C_R which is simple, closed, positively oriented, contains z, and such that f is analytic on and inside this contour. Thus Cauchy's integral formula applies. The added/subtracted contour makes no contribution, and so we have

``` f(z) = 1/(2pi i) int_Gamma f(zeta)/(zeta - z) dzeta
+  1/(2pi i) int_C_R f(zeta)/(zeta - z) dzeta.```
This is almost what we want, but we have an extra term in the expression. If we could show that
`1/(2pi i) int_C_R f(zeta)/(zeta - z) dzeta -> 0 as R -> infinity`
then we could take limits as R tends to infinity, and be done.

To do this, we take advantage of the fact that f(infty) = 0. This means that as R tends to infinity, f(zeta) is going to go to zero; for example, we can make |f(zeta)| < epsilon for any given epsilon, simply by making R sufficiently big. Thus, for sufficiently big R,

` |f(zeta)/(zeta - z)| < epsilon / (R - |z|),`
since |zeta| = R. Since the contour C_R has length 2 pi R, we have
``` |int_C_R f(zeta)/(zeta - z) dzeta| < epsilon 2 pi R/(R-|z|),
so as R tends to infinity we see that the integral can't be any
bigger than 2 pi epsilon.  But epsilon could be anything we wanted,
so the integral has to be zero, which was what we wanted.

Solution B.  (This solution only works when 0 is inside the contour
Gamma).  Make the change of variables zeta = 1/w, dzeta = -dw/w^2.
The expression on the right-hand side becomes
1/(2pi i) int_{1/Gamma} f(1/w) / (1/w - z) (-dw)/w^2,
where 1/Gamma is the image of Gamma under the inversion zeta = 1/w.
After some algebraic manipulation, the expression above can be written
as
1/(2pi i z) int_{1/Gamma} (f(1/w)/w) / (w - 1/z) dw.
Now since f is analytic on and outside Gamma, f(1/w) is analytic on
and inside Gamma, except possibly at 0.  However, since f is analytic
(and is zero) at infinity, f(1/w) is analytic and zero at 0.  Thus
f(1/w)/w has a removable singularity at 0, and so after we
remove the singularity (which doesn't affect the above integral)
we see that f(1/w)/w is analytic everywhere on and inside 1/Gamma.  Thus
the Cauchy Integral Formula applies (note that 1/Gamma is simple closed
and positively oriented), and the above expression is equal to
1/(2pi i z) 2 pi i f(1/(1/z))/(1/z) = f(z),
as desired.

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