On this circle, |z| = R and |Log z| = |ln |z| + i Arg z| = |ln R + i Arg z| <= |ln R| + |i Arg z| <= ln R + pi, where the last inequality is because R > 1 (so |ln R| = ln R) and because -pi < Arg z <= pi (so |i Arg z| = |Arg z| <= pi).
Thus | (Log z)/z^2 | <= (pi + ln R)/R^2 everywhere on C_R. Hence
| / Log z | pi + ln R pi + ln R
| | ----- dz | <= --------- (length of C) = ---------- 2 pi R
| / C z^2 | R^2 R^2
| R |
pi + ln R
= 2 pi --------- .
R
Page 116 #2
/i/2 pi z 1 pi z | i/2 1 i pi/2 pi i
(a) | e dz = -- e | = -- (e - e )
/ i pi | i pi
= (1/pi) (i - (-1) = (1+i)/pi
Parts (b) and (c) are done in exactly the same way; I won't write them all out. The key is being able to recognize that cos(z/2) is the derivative of 2sin(z/2) and that (z-2)^3 is the derivative of (z-2)^4/4.
Page 116 #4
(a) f(z) is analytic except at z=3. The point z=3 lies outside the contour C. This means f(z) is analytic everywhere on and inside C, so the Cauchy-Goursat Theorem guarantees that the contour integral of f(z) over C is zero.
(c) f(z) is analytic except where z^2 + 2z + 2 = 0. Solving this using the quadratic equation gives z = -1 +/- i. These points have modulus sqrt(2) and hence lie outside the contour C. This means f(z) is analytic everywhere on and inside C, so the Cauchy-Goursat Theorem guarantees that the contour integral of f(z) over C is zero.
Page 117 #6 Let D be a circle centred at 2+i, oriented positively, and of small enough radius that it is contained entirely within the rectangle C. For example, you could take D to be of radius 1/2 as illustrated below.
2i +------------------+ 3+2i
| /<-\ |
| D| * 2+i |
| \__/ |
+--------->--------+
0 C 3
Let f(z) = (z-2-i)^n for some power n. Then f(z) is analytic everywhere if n >= 0, and analytic everywhere except at 2+i if n < 0. In either case, it is analytic everywhere between C and D. Therefore,
/ / | f(z) dz = | f(z) dz / C / Dand we know (by the exercise mentioned in the statement of the problem) that this integral is 0 except when n=-1, in which case it is 2 pi i.
Page 117 #8
The integrand is exp(i Log z) which is not analytic everywhere on the contour because it fails to be analytic on the negative real axis. However, everywhere on this contour the integrand is equal to g(z) = exp(i log_(-pi/2, 3pi/2) z) (because these two branches of the logarithm function are equal except in the third quadrant), and g(z) has the advantage of being analytic everywhere on C. We can therefore use the fundamental theorem of calculus for contour integrals to obtain
/ i / | z dz = | g(z) dz = G(1) - G(-1) / C / Cwhere C is the contour in question and G is an antiderivative of g.
To find the antiderivative G, note that we'd expect from our experience with real-variable calculus that G(z) should be some branch of z^(i+1)/(i+1), and it's reasonable to expect it should be the same branch as we chose for g(z). So, let's define G(z) to be this branch (that is, set G(z) = exp[(i+1) log_I z]/(i+1) where I is the interval (-pi/2, 3pi/2)) and verify that G'(z) = g(z):
G'(z) = exp[(i+1) log_I z] (1+i)(1/z) / (1+i)
= exp[i log_I z] exp[ log_I z] (1/z)
= exp[i log_I z] z (1/z)
= g(z).
Our answer is therefore G(1)-G(-1), which equals
G(1)-G(-1) = (exp[(i+1) log_I (1)] - exp[(1+i) log_I (-1)])/(1+i)
= (exp[(i+1)(0)] - exp[(1+i)(i pi)])/(1+i)
= (1 + exp(-pi))/(1+i)
= (1 + e^-pi)/2 (1-i).
Page 118 #9
Split C up into the three pieces C_1, C_2, and C_3 shown:
|
****<**** C3
* | *
* | *
* | *
---*====>====+====>====*-----------
C1 C2
C_1 can be parametrized by z(t) = t (-1 <= t <=0). Here we have f(z(t)) = sqrt(|t|) e^(i pi/2) = i sqrt(|t|) and z'(t) = 1, so
/ /0 /0 1/2 /1 1/2
| f(z) dz = | i sqrt(|t|) dt = i | u (-du) = i | u du
/C1 /-1 /1 /0
3/2 3/2
= i (2/3) [ 1 - 0 ] = 2i/3
(Here I made the
substitution u = |t| = -t, and I am reverting
to the old real-variable calculus meaning of expressions like u^(1/2)
as single-valued real functions).
C_2 can be parametrized by z(t) = t (0 <= t <=1). Here we have f(z(t)) = sqrt(t) e^(i 0/2) = sqrt(t) and z'(t) = 1, so
/ /1 3/2 3/2 | f(z) dz = | sqrt(t) dt = (2/3) [ 1 - 0 ] = 2/3. /C2 /0
Finally, C_3 can be parametrized by z(t) = e^(it), 0 <= t <= pi. Here we have f(z(t)) = sqrt(1) e^(i t/2) and z'(t) = i e^(i t), so
3i pi/2 0
/ /pi it/2 it /pi 3it/2 e - e
| f(z) dz = | e ie dt = | ie dt = i--------------
/C3 /0 /0 (3i/2)
= (-i - 1)/(3/2) = - (i+1) (2/3) = - 2/3 - 2/3 i.
The total integral is therefore 2/3i + 2/3 + (-2/3 - 2/3i) = 0.
The Cauchy-Goursat Theorem does not apply here because f(z) is not analytic at z=0, which is a point on the contour.