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MATH-6640 Spring 2005 COMPLEX VARIABLES AND INTEGRAL TRANSFORMS WITH APPLICATIONS
Solutions to Assignment 2
- Use the Cauchy theorem, the Cauchy integral formula, and suitable extensions to higher derivatives and multiply-connected domains to evaluate the following integrals. Be sure to justify your approach in each case.
(a) (^) ∫
C
z z^3 − 9
dz
where C is the positively oriented rectangle whose sides lie along x = ±5 , y = ±. (b) (^) ∫
C
sin z z^2 (z − 4)
dz,
where C is the positively oriented circle |z| = 2. (c) (^) ∫
C
(z^3 + z + i) sin z z^4 + iz^3 dz,
where C is the positively oriented circle |z| = π. (d) (^) ∫
C
zt z^2 (z + 1)
dz
where C is any positive simple closed contour surrounding |z| = 1.
(a) The integrand is singular at z = zk, i = 1, 2 , 3 , where the zk are the cube roots of 9, given by
zk = 9^1 /^3 ei^2 kπ/^3.
All three roots all lie within C. According to the Cauchy theorem, C can be deformed into three circles Ck, k = 1, 2 , 3 , each of a small radius δ, say, with Ck centered at the singularity zk. Consider the integral around C 1. We can write ∮
C 1
z z^3 − 9
dz =
C 1
z (z − z 2 )(z − z 3 )
z − z 1
dz
= 2πi z 1 (z 1 − z 2 )(z 1 − z 3 )
where the integral has been evaluated by using the CIF (Cauchy Integral Formula). A similar contribution comes from each of the other two contours, and it is easy to see that the three contributions sum to zero. (b) Here we recognize that the integral can be written as ∮
C
f (z) z^2
dz,
where f (z) =
sin z z − 4 is analytic within and on C. Therefore, by using the extended CIF, we have ∮
C
f (z) z^2
dz = 2πif ′(0) = (2πi) (z − 4) cos z − sin z (z − 4)^2
z=
= − 2 πi
πi 2
(c) Upon setting f (z) ≡
(z^3 + z + i) sin z z
and then using partial fractions, the integral can be written as ∫
C
(z^3 + z + i) sin z z^4 + iz^3
dz =
C
f (z)
z
i z^2
z + i
dz
C
f (z) z
dz − i
C
f (z) z^2
dz −
C
f (z) z − i
dz.
Note that f (z) is analytic, save for a removable singularity at z = 0, which is removed by defining f (0) = i. Each term on the RHS above can now be evaluated by using CIF. The result is ∫
C
(z^3 + z + i) sin z z^4 + iz^3 dz = 2πi[f (0) − if ′(0) − f (i)]
= 2πi[i − 0 − i sinh 1] = 2π sinh 1.
(d) Here we use Cauchy’s theorem to deform the contour into two circles of small radii, centered at the singularities z = 0 and z = − 1. Each contribution can then be evaluated by applying CIF. Thus, with f 1 (z; t) = ezt/z^2 and f 2 (z; t) = ezt/(z + 1), we have ∫
C
zt z^2 (z + 1)
dz =
C 1
f 1 (z; t) z + 1
dz +
C 2
f 2 (z, t) z^2
dz
= 2πi[f 1 (−1; t) + f (^2) z (0; t)] = 2πi[e−t^ + t − 1].
- Find the Taylor and Laurent series which represent the function
f (z) =
z^2 − 1 (z + 2)(z + 3)
in the intervals
(a) |z| < 2 , (b) 2 < |z| < 3 , and (c) |z| > 3.
Hint: First express f (z) in terms of partial fractions. The partial-fraction decomposition is
f (z) = z^2 − 1 (z + 2)(z + 3)
z + 2
z + 3
Upon using the binomial expansion
(1 + ξ)−^1 =
∑^ ∞
n=
(−1)nξn^ for |ξ| < 1
one obtains the following.
z + 2
1 2
1 + z 2
n=0(−1) n 2 −n− (^1) zn (^) for|z| < 2 ,
1 z
1 + (^2) z
n=0(−1) n 2 nz−n− (^1) for|z| > 2 ,
(b) f (z) = z/ sin z has a removable singularity at z = 0, simple poles at z = nπ with residues (−1)nnπ, and a nonisolated singularity at infinity. (c) The function f (z) =
z^4 (1 + z^2 )^4
z^4 (z + i)^4 (z − i)^4 has poles of order 4 at z = ±i. The residue at z = i can be obtained either by applying the residue formula or by carrying out a Laurent expansion about z = i. We take the latter approach. We write z = i + t and expand about t = 0. Thus,
f (i + t) =
(i + t)^4 t^4 (2i + t)^4
=
i^4 (1 − it)^4 (2i)^4 t^4 (1 − it/2)^4
=
16 t^4
[1 − 4 it − 6 t^2 + 4it^3 + t^4 ]
×
[
−it 2
−it 2
−it 2
]
The residue is the coefficient of 1/t in the above expansion and is read off as
Res(f (z), z = i) =
[
−i 2
− 4 i
−i 2
−i 2
]
i.
Similarly, Res(f (z), z = −i) =
i.
(d) f (z) = 1/[z(ez^ − 1)] has a pole of order 2 at z = 0 and simple poles at z = 2nπi, n 6 = 0. The residues at the simple poles are − i/(nπ). The residue at the double pole is computed by the formula
Res(f (z), z = 0) = lim z→ 0
d dz
z^2 f (z)
= lim z→ 0
d dz
z ez^ − 1 = −
As in earlier problems, the point at infinity is a limit point of the poles and thus a nonisolated singularity.
- Give the number of branches and locations of the branch points for the functions
(a) cos z^1 /^2 (b) (z + 1)z
(a) Even though z^1 /^2 has a branch point at z = 0 and is double-valued, cos z^1 /^2 is single-valued, since cos(−w) = cos w. In fact, absence of a branch point at z = 0 is confirmed by the Taylor series cos z^1 /^2 = 1 − z 2!
z^2 4!
(b) We have f (z) = ez^ log(z+1). There are branch points at z = −1 and z = ∞, and f (z) is infinitely many-valued.
- Consider
f (z) =
z − 1 z − 1 (a) Select the branch cut to lie along the negative x -axis and define the two branches of f. (b) On each branch, discuss the nature of the point z = 1. In particular, state if it is a singularity, and if so, of what kind. (c) On each branch, expand f in a power series about z = 1 and state whether it is a Taylor series or a Laurent series.
(a) For z^1 /^2 there are branch points at z = 0 and at z = ∞. We choose the cut to be (−∞, 0). Let z = reiθ^ , −π < θ ≤ π. Let w 1 =
reiθ/^2 denote the principal branch of the square root; the second branch being w 2 = −w 1. Then the two branches of f (z) are
f 1 (z) =
reiθ/^2 − 1 reiθ^ − 1
f 2 (z) =
reiθ/^2 − 1 reiθ^ − 1
(b) At z = 1, r = 1 and θ = 0. At this point, f 1 = 0/ 0 and therefore singular. For z 6 = 1, f 1 → 1 /2 as z → 1. Thus, upon defining f 1 (1) = 1/2 the singularity is removed. At z = 1, f 2 = − 2 / 0. The singularity is a simple pole because
lim z→ 1 (z − 1)f 2 (z) = lim z→ 1
reiθ/^2 − 1) = − 2.
(c) In keeping with the above discussion, f 1 should have a Taylor series and f 2 a Laurent series about z = 1. In what follows, z^1 /^2 stands for the principal branch. We set z = 1 + t and expand about t = 0. Then,
f 1 (z) =
z − 1
z − 1)
t [(1 + t)^1 /^2 − 1]
t
t 2
t^2 8
t^3 16
t 4
t^2 8
f 2 (z) = −
z − 1
z + 1)
t
[(1 + t)^1 /^2 + 1]
t
t 2
t^2 8
t^3 16
t
t 8
t^2 16
or z + 1 z − 1
or, z = 13/ 12. Note, according to the definitions above that g 1 (13/12) = 5 while g 2 (13/12) = − 5. Thus the logarithm introduces z = 13/12 as a branch point for f only for the branch g 2 of g. On the other hand, the logarithm forces z = 1 to be a branch point whether g 1 or g 2 is chosen.
