1) The first integral evaluates to 4πi using the Cauchy integral formula applied to circles around z=1 and z=2. 2) The second integral evaluates the 4th derivative of e2z at z=-1 using a formula relating derivatives and contour integrals, giving a value of 24. 3) Both integrals are evaluated quickly using results from complex analysis without direct computation.

1. 1
CAAM 335 Matrix Analysis:
Solutions to HW 7
R
Problem 1 (5+5+5=15 points) (a)Compute the integral C f (z)dz for
f (z) = z C = t 2 + it : t ∈ [0, 2] .
(b)-(c)Verify Cauchy’s theorem for the functions
f (z) = 3z2 + iz − 4,
f (z) = 5 sin(2z),
if C is the square with vertices 1 ± i, −1 ± i.
(a)
Z Z 2 Z 2
f (z)dz = (t 2 − it)(2t + i)dt = 2t 3 + t + i(t 2 − 2t 2 )dt
C 0 0
Z 2
1 1 2 1 2 8
= 2t 3 + t − it 2 dt = t 4 + t 2 |t=0 − i t 3 |t=0 = 10 − i
0 2 2 3 3
(b)-(c) A parametrization of the square with vertices 1 ± i, −1 ± i is given by
C = {z(t) = x(t) + iy(t) : t ∈ [0, 4)} ,
where
 
 −1 + 2t
 t ∈ [0, 1),  −1
 t ∈ [0, 1),
1 t ∈ [1, 2), −1 + 2(t − 1) t ∈ [1, 2),
 
x(t) = y(t) =
 1 − 2(t − 2)
 t ∈ [2, 3),  1
 t ∈ [2, 3),
−1 t ∈ [3, 4), 1 − 2(t − 3) t ∈ [3, 4),
 
Let C j = {z(t) = x(t) + iy(t) : t ∈ [ j − 1, j)} for j = 1, 2, 3, 4. Then
Z 4 Z 4
C
f (z)dz = ∑ f (z)dz = ∑ (g(z( j)) − g(z( j − 1)).
j=1 C j j=1
where g(z) is an anti-derivative of f (z) so that f (z) = g (z).
(b) f (z) = 3z2 + iz − 4 = (z3 + iz2 /2 − 4z) . Hence
Z
f (z)dz = (1 − i)3 + i(1 − i)2 /2 − 4(1 − i) − [(−1 − i)3 + i(−1 − i)2 /2 − 4(−1 − i)] = −2 − 4i
C1
Z
f (z)dz = (1 + i)3 + i(1 + i)2 /2 − 4(1 + i) − [(1 − i)3 + i(1 − i)2 /2 − 4(1 − i)] = 14
C2
Z
f (z)dz = (−1 + i)3 + i(−1 + i)2 /2 − 4(−1 + i) − [(1 + i)3 + i(1 + i)2 /2 − 4(1 + i)] = −2 + 4i
C3
Z
f (z)dz = (−1 − i)3 + i(−1 − i)2 /2 − 4(−1 − i) − [(−1 + i)3 + i(−1 + i)2 /2 − 4(−1 + i)] = 10
C4

2. 2
Clearly, their sum is indeed 0.
(c) f (z) = 5 sin(2z) = (− 5 cos(2z)) . Hence,
2
5 5
Z
f (z)dz = − cos(2(1 − i)) + cos(2(−1 − i)) = 5 sin(2) sinh(2)i
C1 2 2
5 5
Z
f (z)dz = − cos(2(1 + i)) + cos(2(1 − i)) = −5 sin(2) sinh(2)i
C2 2 2
5 5
Z
f (z)dz = − cos(2(−1 + i)) + cos(2(1 + i)) = 5 sin(2) sinh(2)i
C3 2 2
5 5
Z
f (z)dz = − cos(2(−1 − i)) + cos(2(−1 + i)) = −5 sin(2) sinh(2)i
C4 2 2
Again, they sums up to 0.
Problem 2 (5+5+10 =20 points) Find the residues of the following functions at 0:
(z2 + 1)/z, ez /z2 , (2z + 1)/(z(z3 − 5)).
Let us call the three functions f j , j = 1, 2, 3. Then by the residue formula
1 z2 + 1
res( f1 , 0) = lim z = 1,
z→0 (1 − 1)! z
1 d ez
res( f2 , 0) = lim z2 2 = lim ez = 1,
z→0 (2 − 1)! dz z z→0
1 z2 + 1
res( f3 , 0) = lim z = −1/5.
z→0 (1 − 1)! z(z3 − 5)
Problem 3 (5+10=20 points) Let C = 3eit : t ∈ [0, 2π) . Compute the two integrals
sin(πz2 ) + cos(πz2 ) e2z
Z Z
dz, 4
dz.
C (z − 1)(z − 2) C (z + 1)
You may use results from Chapter 8 to arrive quickly at the solution. Be sure to explain your rationale.
Solution 1:
• Let C1 be the circle around a = 1 with radius r = 1/2 and C2 be the circle around a = 2 with radius r = 1/2.
Then,
sin(πz2 ) + cos(πz2 ) sin(πz2 ) + cos(πz2 ) sin(πz2 ) + cos(πz2 )
Z Z Z
dz = dz + dz
C (z − 1)(z − 2) C1 (z − 1)(z − 2) C2 (z − 1)(z − 2)

3. 3
sin(πz2 )+cos(πz2 )
The function f (z) = (z−2) is differentiable on and inside C1 . By the Cauchy Integral Formula,
sin(πz2 ) + cos(πz2 ) sin(π12 ) + cos(π12 )
Z
dz = 2πi f (1) = 2πi = −2πi(sin(π) + cos(π)) = 2πi.
C1 (z − 1)(z − 2) (1 − 2)
sin(πz2 )+cos(πz2 )
The function f (z) = (z−1) is differentiable on and inside C2 . By the Cauchy Integral Formula,
sin(πz2 ) + cos(πz2 ) sin(π22 ) + cos(π22 )
Z
dz = 2πi f (1) = 2πi = 2πi(sin(4π) + cos(4π)) = 2πi.
C2 (z − 1)(z − 2) (2 − 1)
Consequently,
sin(πz2 ) + cos(πz2 )
Z
dz = 2πi + 2πi = 4πi.
C (z − 1)(z − 2)
• We use the equation
dn f n! f (z)
Z
(a) = dz
dan 2πi C (z − a)n+1
with a = −1, n = 3, f (z) = e2z . We can apply this formula for n = 4 and a = −1, since a = −1 is inside C
and f (z) = e2z is differentiable on C and inside C.
f (z) = e2z , f (z) = 2e2z , f (z) = 4e2z , f (z) = 8e2z .
e2z 2πi 2πi 2(−1) 16πi −2 8πi −2
Z
4
dz = f (−1) = 8e = e = e .
C (z + 1) 3! 3! 6 3
Solution 2: Alternatively, we can also use the Residue Theorem to compute both integrals.
• For the first function,
sin(πz2 ) + cos(πz2 ) sin(π) + cos(π)
res(1) = lim (z − 1) = =1
z→1 (z − 1)(z − 2) −1
sin(πz2 ) + cos(πz2 ) sin(4π) + cos(4π)
res(2) = lim (z − 2) = = 1.
z→2 (z − 1)(z − 2) 1
Hence
sin(πz2 ) + cos(πz2 )
Z
dz = 2πi(1 + 1) = 4πi.
C (z − 1)(z − 2)
• For the second function,
1 d3 e2z 1 3 2z 4 −2
res(−1) = lim (z + 1)4 = lim 2 e = e .
z→−1 (4 − 1)! dz3 (z + 1)4 z→−1 6 3
Hence
e2z 8πi −2
Z
4
dz = 2πi res(−1) = e .
C (z + 1) 3

4. 4
Lecture Notes Section 8.5 (P. 93)
Exercise [1] (20 points) Let us confirm the representation (8.7) in the matrix case. More precisely,
if Φ(z) ≡ (zI − B)−1 is the resolvent associated with B then (8.7) states that
h mj
Φ j,k
Φ(z) = ∑ ∑ (z − λ j )k
j=1 k=1
where
1
Z
Φ j,k = Φ(z)(z − λ j )k−1 dz.
2πi Cj
Compute the Φ j,k per (8.15) for the B in (7.13). Confirm that they agree with those appearing in (7.16).
B = [1 0 0; 1 3 0; 0 1 1];
syms z; inv(z*eye(3)-B)
ans =
[ 1/(z-1), 0, 0]
[ 1/(z-1)/(z-3), 1/(z-3), 0]
[ 1/(z-1)ˆ2/(z-3), 1/(z-1)/(z-3), 1/(z-1)]
Thus  
(z − 1)(z − 3) 0 0
1
(zI − B)−1 =  (z − 1) (z − 1)2 0 .
(z − 1)2 (z − 3)
1 (z − 1) (z − 1)(z − 3)
and the eigenvalues of B are λ1 = 1 with multiplicity m1 = 2 and λ2 = 3 with multiplicity m2 = 1.
We use the Cauchy’s Theorem (for differentiable terms) and the Residue Theorem (for terms with singularities)
to compute the following integrals:
 1 
(z−1) 0 0
1 1
Z Z
1 1
Φ1,1 = Φ(z)(z − 1)0 dz = 0  dz

 (z−1)(z−3) (z−3)
2πi C(1,1) 2πi C(1,1)

1 1 1
(z−1)2 (z−3) (z−1)(z−3) (z−1)
1 0 0
   
1
1 0 0
=
 (z−3) 0 0  =  −1 0 0 
 2
1 1
1 −1
4
1
−2 1
(z−3) (z−3) z=1
 
1 0 0
1 1
Z Z
1 (z−1)
Φ1,2 = Φ(z)(z − 1)1 dz = 0  dz

(z−3) (z−3)
2πi 2πi
 
C(1,1) C(1,1) 1 (z−1)
(z−1)(z−3) (z−3) 1
   
0 0 0 0 0 0
= 0 0 0  = 0 0 0 
1
− (z−3) 0 0
z=1
−1
2 0 0

5. 5
 1 
(z−1) 0 0
1 1
Z Z
1 1
Φ2,1 = Φ(z)(z − 3)0 dz = 0  dz
 
(z−1)(z−3) (z−3)
2πi 2πi

C(3,1) C(3,1) 1 1 1
(z−1)2 (z−3) (z−1)(z−3) (z−1)
 
0 0 0
 
0 0 0
1
=

(z−1) 1 0  = 1
1 0 .
 2
1 1 1 1
(z−1)2 (z−1) 0 4 2 0
z=3
Exercise [2] (10 points) Use (8.14) to compute the inverse Laplace transform of 1/(s2 + 2s + 2).
Recall
h
1
Z
(L −1 q)(t) ≡ q(z)ezt dz = ∑ res(λ j )
2πi C j=1
where C is a simple closed curve that encloses the poles of q(z), in this case at z = −1 + i and z = −1 − i.
1 1 ezt
Z
L −1 2 + 2s + 2
(t) = dz = res(−1 + i) + res(−1 − i)
s 2πi C (z − (−1 + i))(z − (−1 − i))
ezt ezt e(−1+i)t e(−1−i)t
= + = +
z − (−1 − i) z − (−1 + i) 2i −2i
z=−1+i z=−1−i
eit − e−it
= e−t = e−t sin(t)
2i
Exercise [3] (20 points) Use the result of the previous exercise to solve, via the Laplace transform, the differential
equation
x (t) + x(t) = e−t sint, x(0) = 0.
Hint: Take the Laplace transform of each side.
Taking the Laplace transform of each side, and using the result from the previous exercise,
L (x (t) + x(t)) = L (e−t sint)
1
sL x − x(0) + L x =
s2 + 2s + 2
1
(s + 1)L x = 2
(s + 2s + 2)
1
Lx =
(s + 1)(s2 + 2s + 2)
Now, we use the inverse Laplace transform to determine x(t).

6. 6
Let C be a sufficiently large circle that encircles the 3 poles at λ3 = −1, λ2 = −1 + i and λ3 = −1 − i.
3
1 1 ezt
Z
L −1 (t) = dz = ∑ res(λ j )
(s + 1)(s2 + 2s + 2) 2πi C (z + 1)(z − (−1 + i))(z − (−1 − i)) j=1
ezt ezt ezt
= + +
z2 + 2z + 2 (z + 1)(z − (−1 − i)) (z + 1)(z − (−1 + i))
z=−1 z=−1+i z=−1−i
e−t e(−1+i)t e(−1−i)t −eit − e−it
= + + = e−t 1 + = e−t (1 − cos(t))
1 (2i)(i) (−2i)(−i) 2
Therefore, the solution to the differential equation is x(t) = e−t (1 − cos(t)).