Solutions to second midterm, 132/1 Winter 2000

Problem 1(a) (5 points): Let gamma_1 be the curve gamma_1(theta) = exp(i theta), 0 <= theta <= 3 pi.  Compute the integral of zdz along gamma_1, where z is the conjugate of z.

Answer: Making the change of variables

z = exp(i theta)
dz = i exp(i theta) d theta

and observing that

z = exp(- i theta)
the integral becomes that of

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:

• Ignoring the conjugation, and writing z as exp(i theta) instead of  exp(- i theta).  (Or ignoring the z term altogether).
• Using the fundamental theorem of calculus, and trying to press z^2/2 or z^2/2 into service as an anti-derivative for z.  (Actually z does not have an anti-derivative, because it is not analytic, and only analytic functions have anti-derivatives).
• It is possible to "guess" the answer as 3 pi i, based on reasoning by analogy with similar examples.  However, this type of reasoning can be dangerous.  For instance, suppose we are integrating z on the curve z = 2 exp(i theta), 0 <= theta <= 2 pi.  Can you guess the answer by inspection?  Make a guess, and then check your answer.

Problem 1(b) (10 points):  Let gamma_2 be the line segment from -1 to -1-i.  Compute the integral of dz/z on gamma_2.

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:

• Interchanging the -1 and -1-i.  If F is an anti-derivative of 1/z, the integral from -1 to -1-i of dz/z is F(-1-i) - F(-1), not F(-1) - F(-1-i).
• Using the multi-valued logarithm instead of a branch of a logarithm.  Taken to extremes, this approach would not give a single answer to an integral, but the answer would have a term like + 2 k pi i in it.  This is absurd - this integral can only have one specific value, not a range of values.
• Evaluating the logarithm incorrectly - either getting a completely incorrect imaginary part, or an imaginary part which did not lie in the specified interval (in this case, (0,2 pi]).
• Using ln|z| as an anti-derivative for 1/z.  This is simply not true, even though ln|x| is an anti-derivative for 1/x.  ln|z| isn't even an analytic function.  The only anti-derivatives of 1/z are the branches of the complex logarithm.
• A variant of the previous error was to first make a change of variables such as z = -1-it, and then using ln|1+it| (or something like this) as an anti-derivative of 1/(1+it).  Again, when integrating a complex reciprocal, you need to use the complex logarithm, not the real logarithm.
• Combining logarithms by the rule Log(z/w) = Log(z) - Log(w) or Log(zw) = Log(z) + Log(w).  This rule works sometimes, but is dangerous; you can sometimes be off by 2 pi i.  (Why?)
If gamma_2 was a more complicated curve, e.g. a curve that wound around the origin two-and-a-half times, then one could not easily use the above method.  However, one could break the curve into shorter pieces and use the Fundamental theorem on each piece (possibly using a different branch of the logarithm for each piece).  If the curve was closed, you could also try to use Cauchy's integral formula.

Problem 2(a) (10 points)

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.

Problem 3(a) (10 points): Let gamma be the circle of radius 2 centered at 2i, traversed once anti-clockwise.  Compute the integral of 2 dz / (z^2 + 4)
on gamma.

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

[2/(z-2i)]  /  (z+2i)

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:

• Many people wrote 2/(z+2i)(z-2i) as the sum of [2/(z+2i)]/(z-2i) and [2/(z-2i)]/(z+2i).  This is incorrect.  The three expressions are equal; the first is not equal to the sum of the other two.
• Some people factorized (z^2+4) incorrectly as (z+2)(z-2) or (z+2i)^2.
Problem 3(b) (10 points): find a parameterization of gamma.

One possibility is

gamma(t) = 2i + 2 exp(it)    0 <= t <= 2pi

There are several other correct answers.

Problem 4(a) (10 points): Let Gamma be the circle |z|=2 traversed once in the negative direction.  Compute the integral of exp(2z)/(1-z)^2 dz on Gamma.

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:

• Neglecting to change the sign because of the negative orientation of Gamma.
Problem 4(b) (10 points).  Compute the integral of sin(z)/(exp(z) (z-3)) on the same contour Gamma.

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.

Problem 5(a) (6 points): Find the first three terms of the Taylor expansion of 1/(1+z) around z = 0.

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.