Answer: Making the change of variables
z = exp(i theta)
dz = i exp(i theta) d theta
and observing that
exp(-i theta) i exp(i theta) d theta
from 0 to 3pi. The exponentials cancel, so you are just integrating i dtheta from 0 to 3 pi, which is 3 pi i.
Some common errors:
Answer: We shall apply the fundamental theorem of calculus. We
want an anti-derivative of dz/z which is analytic on gamma_2. The
principal branch Log(z) won't work, because it's not analytic
on the negative real line, and gamma_2 just touches this line. However,
a branch such as Log_(0,2 pi] (z) is analytic
on gamma_2 (it is only non-analytic at 0 and the positive real axis,
which is not a problem). So by the Fundamental theorem of calculus
the answer is
Log_(0,2 pi](-1-i) - Log_(0,2 pi](-1).
Since
Log_(0,2 pi](-1-i) = ln(sqrt(2)) + 5 pi i / 4
and
Log_(0,2 pi](-1) = ln(1) + pi i = pi i
we thus get a final answer of ln sqrt(2) + pi i / 4.
Some common errors:
Answer: gamma_1 is the line segment from 0 to 2; gamma_2 is the line segment from 0 to 2i; gamma_3 is the line segment from 2 to 2i.
Problem 2(b) (10 points) Suppose f is an entire function such that the integral of f on gamma_1 is 5 pi i and the integral of f on gamma_2 is 3 pi i. What possible values can the integral of f on gamma_3 take? Explain.
The contour gamma_1 + gamma_3 + -gamma_2 is a simple closed contour. Since f is analytic, it is analytic on and inside this contour, so the integral along this contour is 0. In other words,
[integral on gamma_1] + [integral on gamma_3] - [integral on gamma_2] = 0
so
5 pi i + [integral on gamma_3] - 3 pi i = 0
so the integral on gamma_3 is -2pi i.
Factorize 2/(z^2+4) as 2/(z+2i)(z-2i). At this point one can use partial fractions, but the quickest way is to rearrange the integrand. One could write the integrand as
but this isn't very helpful, because the numerator is not analytic on and inside the contour (it has a singularity at 2i), and the singularity in the denominator is at -2i, which is outside the contour. So a better way to write the integrand is as
[2/(z+2i)] / (z-2i).
The integrand is now analytic on and inside the contour (-2i is outside the contour), and the singularity in the denominator is at 2i, which is inside the contour. Since the contour is simple, closed, and anti-clockwise, we can now apply Cauchy's Integral Formula to compute the integral as
2 pi i [ 2 / (2i + 2i) ] = pi.
Common errors:
One possibility is
gamma(t) = 2i + 2 exp(it) 0 <= t <= 2pi
There are several other correct answers.
Answer: We use the generalized Cauchy Integral Formula. First we rewrite the integral as
exp(2z)/(z-1)^2.
The function f(z) = exp(2z) is entire, so is analytic on and inside the contour, and the singularity 1 is inside Gamma. If Gamma was anti-clockwise, the generalized Cauchy integral formula would then give an answer of
2 pi i f'(1)
but since Gamma is clockwise, the correct answer is
- 2 pi i f'(1).
Since f'(z) = 2 exp(2z), this simplifies to
- 4 pi i e^2.
Common errors:
The integrand is analytic at every point except at 3 (note that exp(z) is never zero). In particular, it is analytic on and inside the contour Gamma, so by Cauchy's theorem the integral is zero.
One can either modify the geometric series formula 1/(1-z) = 1 + z + z^2 + ..., or use Taylor's formula directly. The answer is
1/(1+z) = 1 - z + z^2 - ... for all |z| < 1.
Problem 5(b) (6 points): Find the first three terms of the Taylor expansion of 1/(1+z) around z=-2.
One can either write 1/(1+z) = -1/(1 - (z+2)) and then apply the geometric series formula, or instead use Taylor's formula. The answer is
1/(1+z) = -1 - (z+2) - (z+2)^2 - ... for all |z+2| < 1.
Problem 5(c) (8 points): Find the first three terms of the Taylor expansion of 1/(1+z)^3 around z=0.
One can differentiate the answer to 5(a) twice, or apply Taylor's formula. (One can also try to cube the answer to 5(a) but this is not very efficient). The answer is
1/(1+z)^3 = 1 - 3z + 6z^2 - ... for all |z| < 1.
In retrospect, I should not have set up this question, given that it can be attacked most efficiently by a brute force application of Taylor's formula.