Section 4.1

Question 1: See solutions in textbook. Other parameterizations are possible, but you should get something which looks about the same as the answers in the textbook.

Question 8: We break up the contour into two curves gamma_1 and gamma_2:

gamma_1(t) = -2+2i + (1-2i)t, 0 <= t <= 1

gamma_2(t) = e^{-it}, -pi <= t <= 0the question asks us to combine these two paramterizations. One way is to shift the second curve (say) from the interval [-\pi,0] to the interval [1, 1+\pi] by the change of variables s = t + pi + i:

gamma_2(s) = e^{-i(s-pi-1)}, 1 <= s <= 1+piand combine the two as

Gamma(t) = -2+2i + (1-2i)t, 0 <= t <= 1

Gamma(t) = e^{-i(t-pi-1)}, 1 <= t <= 1+piThe reverse contour could be given by

-Gamma(t) = e^{-i(-t-pi-1)}, -1-pi <= t <= -1

-Gamma(t) = -2+2i + (1-2i)(-t), -1 <= t <= 0Other parameterizations are possible. In practice it is not particularly advantageous to obtain a single paramterization for a complicated contour, as you will most likely need to break it back into components in any event. However occasionally it is useful to have a parameterization by a single function instead of a number of unrelated functions.

Question 10: (a) Take gamma(t) = z_1 + t(z_2 - z_1), 0 <= t <= 1; the length formula gives

l(gamma) = int_0^1 |gamma'(t)| dt = int_0^1 |z_2 - z_1| dt = |z_2 - z_1|.Here I use int to represent the integration symbol.

(b) Take gamma(t) = z_0 + re^{it}, 0 <= t <= 2pi; the length formula gives

l(gamma) = int_0^2pi |gamma'(t)| dt = int_0^2pi |rie^{it}| dt = int_0^2pi r = 2pi rsince i and e^{it} have magnitude 1.

Section 4.2

Question 3(b): Using our real integration knowledge as a guide, we guess that the antiderivative of (1+i) cos(it) is (1+i) sin(it)/i; we check this by differentiation. So the integral is just this antiderivative evaluated at 0 and at -2; since sin(0) = 0 and sin(-2i) = i sinh(-2) = -2 sinh(i), we get -2(1+i)sinh(2) for the integral.

You are perfectly justified in using real integration techniques to **guess**
the antiderivative of a function, but before you apply it to a complex integral
you should actually check by differentiation that you genuinely have an
anti-derivative. The classic example is when integrating 1/x; using the real
integral as a guide you might guess ln|z| is an anti-derivative, but it is not;
it is not even complex differentiable.

Question 5. Each term in the integral is a special case of Example 2, so the answer is 6 * 0 + 2 * 2 pi i + 1 * 0 - 3 * 0 = 4 pi i.

Question 7. If we use the parameterization z(t) = (1+2i)t for 0

Question 11. Simplest way is by the fundamental theorem of calculus: all three integrals are equal to z^2 + z evaluated at 1 and -i, i.e. (1^2 - 1) - ((-i)^2 - i) = i + 1.

Question 13. On this curve the first term is 1, so we may replace the integrand by (1-z). Now the answer is just z-z^2/2 evaluated at the endpoints -i and 2+i, i.e.

(-i + 1/2) - (2 - i - (2-i)^2 / 2) = -2i.

Section 4.3

Question 1. (b) e^z is an anti-derivative, so the answer is e^z evaluated at -1 and 1, i.e. (e^{-1} - e).

(d) cot z is an anti-derivative of cosec^2 except on integer multiples of 2pi, so the FToC applies and we get 0.

(f) The first step is to find an anti-derivative for e^z cos z. One way to do this is to find the real anti-derivative of e^x cos x and use that as a guess for the complex anti-derivative of e^z cos z; this is usually a good way to find anti-derivatives. Or we can take advantage of the formula cos z = (e^{iz} + e^{-iz})/2. Once we have our anti-derivative (in this case, it will be (e^z cos z + e^z sin z)/2), we evaluate it at i and pi to get (e^pi + e^i (cosh 1 + i sinh 1)/2.

(g) We have to find the anti-derivative of exp(1/2 Log z), the principal branch of z^{1/2}. The real anti-derivative of x^{1/2} is 2/3 z^{3/2}, so one would expect the anti-derivative of exp(1/2 Log z) to be something similar, e.g. the principal branch of 2/3 z^{3/2}, which is 2/3 exp(3/2 Log z). Let us now check this guess by differentiation:

d/dz 2/3 exp(3/2 Log z) = 2/3 3/2 exp(3/2 Log z) (d/dz Log z) = exp(3/2 Log z)/z

= exp(3/2 Log z) / exp(Log z) = exp(1/2 Log z),as desired. Note that this function is only an anti-derivative away from the negative real axis, because Log z is not differentiable there. So the integral is just 2/3 exp(3/2 Log z) evaluated at i and pi, which is 2/3( -sqrt(2)/2 + sqrt(2)i/2 - pi^{3/2} ).

Question 2. Every polynomial has an anti-derivative on the entire complex plane, so the integral of a polynomial on any closed contour is zero.

Question 5. If 1/z had an anti-derivative in this domain, then the integral of 1/z around any closed contour in this domain would be zero. But this is not true, as for example the integral of 1/z on the positively oriented unit circle is 2 pi i, not 0.