Problems and Solutions in - Mathematics - University of Hawaii

2.2 1991 November 21 2 COMPLEX ANALYSIS
where a and each root λn are complex constants. Prove that if P has only real coefficients, then P has a factorization
K
M
P (z) = a (z − rk) (z 2 − bmz + cm),
k=1
where a and each rk, bm, cm are real constants.
m=1
3. Use complex residue methods to compute the integral
π
1
5 + 3 cos θ dθ.
Solution: Let I = π
0
For z = e iθ ,
0
1
5+3 cos θ dθ. Note that cos θ is an even function (i.e., cos(−θ) = cos θ), so
2I =
and dz = ieiθdθ, from which it follows that
2I =
π
−π
cos θ = eiθ + e −iθ
2
= 1
i
= 2
3i
|z|=1
|z|=1
|z|=1
1
5 + 3 cos θ dθ.
= 1 1
(z +
2 z ),
1
5 + 3 1
2 (z + z )
dz
iz
dz
5z + 3
2 (z2 + 1)
dz
z2 + 10
3 z + 1.
Let p(z) = z 2 + 10
3 z + 1. Then the roots of p(z) are z1 = −1/3 and z2 = −3. Only z1 = −1/3 is inside the circle
|z| = 1, so the residue theorem implies
Now,
which implies
Therefore,
2I = 2
1
· 2πi · Res , z1 .
3i p(z)
1
Res , z1 = lim
p(z) z→z1
1
p(z) =
1
(z − z1)(z − z2) ,
1
=
z − z2
1
− 1
3
=
3 − (−3) 8 .
2I = 2 3 π
· 2πi · =
3i 8 2 ,
so I = π
4 . ⊓⊔
44