MATH 3160: APPLIED COMPLEX VARIABLES
TEST #2 (VERSION B)
1 . Find
Z
| z | =1 z cos( z ) dz.
sin( z )
Solution: The singularities of 1 / sin( z ) are at z = πN for N any integer.
The only one inside the circle | z | = 1 is at z = 0. But there, we can see that z/ sin( z ) has a removable singularity. Therefore the integral vanishes.
2 . Consider the integral
I =
Z
∞
0 x
α
(1 + x 2 )
2 dx.
(a) Explain how to set up an integration along a branch cut for this problem. Include a picture of the contour you want to use.
(b) Show that as long as − 1 < α < 3 and α = 1 and α = 2 then the integral I is finite.
Solution:
(a) We choose to cut a branch for z α along the positive real axis. The contour we integrate on is shown in figure 1.
Date : July 30, 2001.
Figure 1.
Integrating along the branch, then around a large circle, then backwards along the branch, and then around a small circle.
1

2 MATH 3160: APPLIED COMPLEX VARIABLES TEST #2 (VERSION B)
(b) The integral of the function f ( z ) = z α
(1 + z 2 )
2 around a circle of radius r (starting at angle θ = 0 and going to θ = 2 π ) is
Z Z
2 π f ( z ) dz =
0 r
α +1 e i ( α +1) θ
1 + 2 r 2 e 2 iθ + r 4 e 4 iθ i dθ which for r large behaves like r α − 3 while for r small it behaves like r α +1 . This makes the integrals around both circles in our contour go to zero for α in the range − 1 < α < 3, as the small circle gets very small and the large circle very large. But then the integral we are looking for is just a multiple of the residue inside. In fact at θ = 0 the integral along the real line of f ( z ) is precisely the integral I we started with, while for θ = 2 π it is
Z
0
∞ x α e 2 πiα
(1 + x 2 )
2 dx = − e
2 πiα
I.
So the total contour integral (for any sufficiently small and sufficiently large circles) is
Z contour f ( z ) dz = 1 − e
2 πiα
I.
The trouble at α = 1 or α = 2 is that this equation doesn’t allow us to solve for I in terms of the contour integral, because the right hand side is just zero. The value of the integral is
(1 − a ) π
I =
4 cos ( aπ/ 2) but you don’t have to find it for this problem.
3 .
(a) Calculate the Laurent expansion about z = 0 of e z sin z up to z 2
(b) Calculate terms.
e z
Res z =0 sin z
Solution:
(a) e z sin z
=
1 z
+ 1 +
2
3 z +
1
3
You only need the first four terms.
z
2
+ O z
3
(b) e z
Res z =0 sin z
= 1

MATH 3160: APPLIED COMPLEX VARIABLES TEST #2 (VERSION B) 3
4 . Calculate the integral
Z
| z | =1 z 3 dz cosh z
.
Solution: The only singular point inside the circle is at z = 0, since the cosh z vanishes only at odd multiples of πi/ 2, which are not in the circle.
The residue at z = 0 is z 3
1
Res z =0 cosh z
Therefore the value of the integral is
= −
1
2
Z dz
| z | =1 z 3 cosh z
= 2 πi Res z =0
1 z 3 cosh z
= − πi
5 . Calculate the integral
Z
| z | =1 z dz cosh (1 /z )
.
Solution: The function z cosh (1 /z ) has poles at
1 z =
(2 N − 1) πi for any integer N ; these poles all lie inside the circle | z | = 1, and accumulate at the origin. So the origin is not an isolated singular point, and we can’t use a residue calculation. Instead we will calculate the residue at infinity.
Change variable: w = z
− 1 so that dw = − z
− 2 dz.
Also, travelling around the circle | z | = 1 counterclockwise makes w go around the circle | w | = 1 clockwise. This gives
Z z dz
| z | =1 cosh (1 /z )
= −
Z
| w | =1 dw w 3 cosh w which you know from the previous problem.