MATH 20101 Complex Analysis
Solution Sheet 6
1.(a) We have
1 1
f (z) = =
z2 5z + 6 (z 2)(z 3)
so the integrand has simple poles at z = 2 and z = 3 both inside C4 . Hence
Z
1
2 5z + 6
dz = 2iRes(f, 2) + 2iRes(f, 3).
C4 z
We have
(z 2) 1
Res(f, 2) = lim = lim = 1
z2 (z 2)(z 3) z2 z 3
and
(z 3) 1
Res(f, 3) = lim = lim = 1.
z3 (z 2)(z 3) z3 z 2
Thus the value of the integral is
2i + 2i = 0.
(b) Now we have the same function integrated over the smaller circle C5/2 . This time only
the pole z = 2 lies inside. So
Z
1
2 5z + 6
dz = 2iRes(f, 2) = 2i.
C5/2 z
RR R
2. Remember that the real integral R f (x)dx is equal to the complex integral [R,R] f ,
where [R, R] is the line segment from R to R.
RR 1
(a) First note that the integral converges, so it is equal to its principal value limR R 1+x 2 dx.
For R > 1, let SR denote the semi-circular path
SR (t) = Reit , 0 t ,
and let R = [R, R] + SR (a D-shaped contour). [Draw a picture!]
The integrand f (z) = (z 2 + 1)1 = (z i)1 (z + i)1 has simple poles at z = i and z = i;
only z = i lies inside R (for R > 1). Hence
(z i)
Z Z Z
1 2i
f+ f= f = 2iRes(f, i) = 2i lim = 2i lim = = .
[R,R] SR R zi (z i)(z + i) zi z + i 2i
Now we show that the integral over SR tends to zero, as R . O n SR , |z 2 + 1| |z 2 | 1 =
R2 1, so (using the Estimation Lemma)
Z
R
f 2 0,
SR R 1
1
as R . Therefore Z Z
1
= lim f = .
x2 + 1 R [R,R]
(One can also do this by recalling that (x2 + 1)1 has antiderivative arctan x.)
R R e2ix
(b) Here the integral also converges, so it is equal to its principal value limR R 1+x 2 dx.
2iz 2 1 2iz 1 1
The integrand g(z) = e (z + 1) = e (z i) (z + i) has simple poles at z = i and
z = i; only z = i lies inside R (for R > 1). Hence
(z i)e2iz e2iz 2ie2
Z Z Z
g+ g= g = 2iRes(g, i) = 2i lim = 2i lim = = e2 .
[R,R] SR R zi (z i)(z + i) zi z + i 2i
As in (a), we show that the integral over SR tends to zero, as R . For z = x + iy SR ,
|e2iz | = e2y 1 and (from (a)) |z 2 + 1| R2 1. So, by the Estimation Lemma,
Z
R
g 2 0,
SR R 1
as R . Therefore
e2ix
Z Z
= lim g = e2 .
x2 + 1 R [R,R]
R e2ix
Since the integral x2 +1 dx converges, its real and imaginary parts both converge. Hence
Z Z
cos 2x sin 2x
= Re(e2 ) = e2 and = Im(e2 ) = 0.
x2 + 1 x2 + 1
(If e2ix were replaced by e2ix we could not use the semi-circle SR since for z = x + iy SR ,
we would only have the bound |e2iz | = e2y e2R . To get around this, use the negative
semi-circle
Reit , 0 t
and the other pole z = i.)
3. We will use the same notation SR , R as in question 2. Both the integrals converge, so
they will be equal to their principal values.
(a) Write f (z) = (z 2 + 1)1 (z 2 + 3)1 . This has simple poles at z = i, z = i, z = 3i and
z = 3i. If R > 3 then z = i and z = 3i are contained inside the D-shaped contour R .
Hence Z Z Z
f+ f= f = 2iRes(f, i) + 2iRes(f, 3i).
[R,R] SR R
We have
zi 1 1 1
Res(f, i) = lim = lim = =
zi (z 2 + 1)(z 2 + 3) zi (z + i)(z 2 + 3) 2i(1 + 3) 4i
and
z 3i 1 1 1
Res(f, 3i) = lim
2 + 1)(z 2 + 3)
= lim
= = .
z 3i (z 2
z 3i (z + 1)(z + 3i) (3 + 1)2 3i 4 3i
2
Substituting, we get
Z Z
2i 2i 1
f+ f= = = 1 .
[R,R] SR 4i 4 3i 2 2 3 2 3
Now we show that the integral over SR tends to zero, as R . For z on SR , |(z 2 + 1)(z 2 +
3)| (|z 2 | 1)(|z 2 | 3) = (R2 1)(R2 3). So, by the Estimation Lemma,
Z
R
f 0,
SR (R 1)(R2 3)
2
as R . Therefore
Z Z
1 1
dx = lim f= 1 .
(x + 1)(x2 + 3)
2 R [R,R] 2 3
(b) Note that
28 + 11x2 + x4 = (x2 + 4)(x2 + 7).
2 + 4)1 (z 2 + 7)1 . This has simple poles at z = 2i, z = 2i, z =
So write
g(z) = (z 7i and
z = 7i. If R > 7 then z = 2i and z = 7i are contained inside the D-shaped contour
R . Hence Z Z Z
g+ g= g = 2iRes(g, 2i) + 2iRes(g, 7i).
[R,R] SR R
We have
z 2i 1 1 1
Res(g, 2i) = lim = lim = =
z2i (z 2 2 2
+ 4)(z + 7) z2i (z + 2i)(z + 7) 4i(4 + 7) 12i
and
z 7i 1 1 1
Res(f, 7i) = lim
2 2
= lim
= = .
z 7i (z + 4)(z + 7) 2
z 7i (z + 4)(z + 7i) (7 + 4)2 7i 6 7i
Substituting, we get
Z Z
2i 2i 1 1
f+ f= = = .
[R,R] SR 12i 6 7i 6 3 7 3 2 7
Now we show that the integral over SR tends to zero, as R . For z on SR , |(z 2 + 4)(z 2 +
7)| (|z 2 | 4)(|z 2 | 7) = (R2 4)(R2 7). So, by the Estimation Lemma,
Z
R
g 2 4)(R2 7)
0,
SR (R
as R . Therefore
Z Z
1 1 1
dx = lim g= .
28 + 11x2 + x4 R [R,R] 3 2 7
3
4. We will use C to denote the unit circle
C(t) = eit , 0 t 2.
(a) Substitute z = eit . Then dt = dz/iz, cos t = (z + z 1 )/2 and [0, 2] transforms into C.
Hence
(z + z 1 )3 3(z + z 1 )2 dz
Z 2 Z
3 2
2 cos t + 3 cos dt = + .
0 C 4 4 iz
We have
(z + z 1 )3 z 3 + 3z + 3z 1 + z 3
=
4 4
and
3(z + z 1 )2 3z 2 + 6 + 3z 2
=
4 4
so Z 2
3 2z z 2
Z
3 2 1 1 3 3 3
2 cos t + 3 cos dt = + + + + + + dz.
0 i C 4z 4 4z 3 4z 2 2z 4 4 4
We see that the integrand has a pole of order 4 at z = 0, which is inside C, and no other
poles. We can immediately read off the residue at z = 0 as the coefficient of 1/z, namely 3/2.
Hence Z 2
1 3
2 cos3 t + 3 cos2 dt = 2i = 3.
0 i 2
(b) As before, substitute z = eit . Then dt = dz/iz, cos t = (z + z 1 )/2, and [0, 2] transforms
into C. Hence
Z 2 Z Z
1 1 dz 1 4z
2
dt = 1
= dz.
0 1 + cos t C 1 + (z + z )2/4 iz i C z + 6z 2 + 1
4
Try to factorize z 4 + 6z 2 + 1 = (z 2 + )(z 2 + ). Then + = 6 and = 1 so we can take
= 3 2 2, = 3 + 2 2.
Note that |3 + 2 2| > 1 and so |3 2 2| < 1. Thus we see that
4z
f (z) =
+ 6z 2 + 1z4
p p
has simple poles inside C located at z = i = i 3 2 2 and z = i = i 3 2 2.
We have
(z i )4z 4i 2
Res(f, i ) = lim 2 2
= =
zi (z + )(z + ) 2i ( )
and
(z + i )4z 4i 2
Res(f, i ) = lim 2 2
= = .
zi (z + )(z + ) 2i ( )
Hence
2
Z
1 1 2 2 8 8
dt = 2i + 2i = = = 2.
0 1 + cos2 t i 4 2
