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1
Final Exam M601 (Notes, books, and calculators are not authorized) Show all your work in the
blank space you are given on the exam sheet. Always justify your answer. Answers with no
justification will not be graded.
Question 1: Is the function z 7−→ f (z) = e−y ((1 + x) cos(x) − y sin(x)) + ie−y ((1 + x) sin(x) +
y cos(x)), where z = x + iy, holomorphic over C? Justify your answer.
The real part and imaginary part of f are clearly differentiable over C. We just have to verify
whether the Cauchy-Riemann relations hold. Let P (x, y) = e−y ((1 + x) cos(x) − y sin(x)) and
Q(x, y) = e−y ((1 + x) sin(x) + y cos(x))
∂x P (x, y) = e−y (−(1 + x) sin(x) + cos(x) − y cos(x))
∂y Q(x, y) = e−y (−(1 + x) sin(x) − y cos(x) + cos(x))
and
∂y P (x, y) = e−y (−(1 + x) cos(x) + y sin(x) − sin(x))
∂x Q(x, y) = e−y ((1 + x) cos(x) + sin(x) − y sin(x))
Clearly ∂x P (x, y) = ∂y Q(x, y) and ∂y P (x, y) = −∂x Q(x, y); as a consequence, f is holomorphic.
R
Question 2: Evaluate C 3z 2 dz along the curve C = γ([0, 1]) where γ(t) = t.
Answer 1: Since dz = γ 0 (t)dt = dt, we have
Z
Z
2
z dz =
C
1
3t2 dt = t3 |10 = 1.
0
Answer 2: Observe that 3z 2 = f 0 (z) where f (z) = z 3 ; as a result
Z
Z
2
3z dz =
f 0 (z)dz = f (1) − f (0) = 1
C
C
2
Final Exam, December 13, 2011
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Question 3: Evaluate C 2zdz along the curve C = γ([0, 3]) where γ(t) = it for 0 ≤ t ≤ 1,
γ(t) = t − 1 + i(1 + sin(50π(t − 1))) for 1 ≤ t ≤ 2, and γ(t) = 1 + i(3 − t) for 2 ≤ t ≤ 3.
Observe that 2z = f 0 (z) where f (z) = z 2 and f is clearly holomorphic. As a result
Z
Z
2zdz =
f 0 (z)dz = f (γ(3)) − f (γ(0)) = f (1) − f (0) = 1.
C
C
Question 4: Using the notation z = reiθ , where r ≥ 0, θ ∈ [0, 2π), consider the function log(z)
defined as follows log(z) := log(r) + iθ. Prove that log(z) is not continuous across the axis
{y = 0, x > 0}.
Let z0 = r be an arbitrary number on the positive real axis {y = 0, x > 0}. Let us set z+ () = rei ,
z− () = rei(2π−) with > 0. It is clear that
lim z+ () = z0 = rei2π = lim z− (),
→0
→0
i.e., both z+ () and z− () converge to z0 as goes to zero. If the complex square root function
defined above was continuous, the two quantities log(z+ ) and log(z− ) should converge to the same
value as goes to zero. But using the above definition of log(z) we have
lim log(z+ ) = lim log(r) + i = log(r)
→0
→0
and
lim log(z− ) = lim log(r) + i(2π − ) = log(r) + 2π.
→0
→0
In conclusion lim→0 log(z+ ) 6= lim→0 log(z− ), thereby proving that log(z) is not continuous across
the axis {y = 0, x > 0}.
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3
Question 5: Let f (z) = (2z 2 + 3z + 1)z −3 . Compute
3π/5 centered at 21 traversed counterclockwise.
R
C
f (z)dz where C is the circle of radius
f (z) = 2z −1 + 3z −2 + z −3 is holomorphic except at 0 and zero is a pole of third order. The residue
of f at zero is Res(f, 0) = 2 The open domain limited by the circle of radius 3π/5 centered at 12
contains the pole of f ; as a result
Z
f (z)dz = 2iπRes(f, 0) = 4iπ.
C
R 2π
iθ
Question 6: Compute the integral I = 0 5−4dθ
cos(θ) . (Hint: Use the parameterization z = e
and observe that cos(θ) = 21 (z + z1 ), dz = izdθ. Conclude with the residue Theorem.)
We use the hint. We set γ(θ) = eiθ and C = γ(0, 2π), and observe that dz := γ 0 (θ)dθ = ieiθ dθ =
1
dz. Note that C is the unit circle centered at 0. We have
izdθ, i.e., dθ = iz
Z
2π
I=
0
Let us define f (z) =
dθ
=
5 − 4 cos(θ)
−1
i(2z 2 −5z+2)
=
Z
C
dz
=
iz(5 − 2(z + z −1 ))
−1
.
2i(z− 12 )(z−2)
Z
C
dz
i(5z − 2z 2 − 2)
This function is holomorphic in C expect at
and 2. We apply the residue theorem after observing that only the pole at
1
I = 2iπRes(f, ).
2
with
1
1
1
−1
=
Res(f, ) = lim1 (z − )f (z) =
2
2
3i
2i( 12 − 2)
z→ 2
Finally,
I = 2π
1
2π
=
3i
3
1
2
is in the unit circle:
1
2
4
Final Exam, December 13, 2011
Question 7: Is the matrix A =
2
7
0
diagonalizable. Be very accurate when answering.
2
There is a unique eigenvalue λ = 2.
Answer 1: If A was diagonalizable it could be written as follows
A = P DP −1
where D = 2I. That would mean that
A = 2P IP −1 = 2P P −1 = 2I,
which is wrong. In conclusion A is not diagonalizable.
Answer 2: Let v = (x, y) be an eigenvector, then
0 0
x
(A − 2I)v =
=0
7 0
y
This means v = (0, y) = y(0, 1), i.e., the eigenspace associated with the only eigenvalue 2, is
one-dimensional E2 = span{(0, 1)}. It is not possible a set of two independent eigenvectors. This
means that A is not diagonalizable.


1 2 0
Question 8: Consider the matrix 0 3 0. The eigenvalues of A are 1, 2, 3. Is the matrix
2 −4 2
A diagonalizable? Be very accurate in your statements.
The 3×3 matrix A has three distinct eigenvalues, hence A is diagonalizable. (Theorem from class)
Question 9: Let du
dt = F (u) be five-dimensional nonlinear dynamical system. Assume that
u0 ∈ R5 solves F (u0 ) = 0 and assume that the dynamics in the neighborhood of u0 is described
√1 , − 1 + 3i,
by dz
dt = Az where A is 5 × 5 real-valued matrix with eigenvalues −2 + 4i, −2 − 4i,
3
10
1
− 3 − 3i. Is u0 a stable equilibrium solution? Justify your answer.
The real part of the eigenvalue
√1
10
is positive. As a result u0 is a unstable equilibrium point.
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5
Question 10: Let A be a n×n complex-valued Hermitian matrix (i.e., AT = A), n ≥ 2. We
define the complex inner product hv, wi := v T w and associated norm kvk2 = v T v.
(a) Show that hv2 , Av1 i = hAv2 , v1 i, for all vectors v1 , v2 in Cn . Give all the details.
hv2 , Av1 i = v2T Av1 = v2T Av1 = v2T AT v1 = (Av2 )T v1 = hAv2 , v1 i.
(b) Prove that the eigenvalues of A are real. Be very accurate in your statements.
Let (v, λ) be an eigenpair. Then
hAv, vi = (Av)T v = λv T v = λkvk2
and
hv, Avi = v T Av = λv T λv = λkvk2 .
From (a) we know that hAv, vi = hv, Avi; as a result
λkvk2 = λkvk2 ,
which implies λ = λ since v 6= 0.
(c) Let λ1 and λ2 be two different eigenvalues of A and let v1 , v2 be two associated eigenvectors.
Prove that v1 and v2 are orthogonal with respect to the complex inner product hv, wi := v T w.
We have
hv2 , Av1 i = hv2 , λ1 v1 i = v2T (λ1 v1 ) = λ1 v2T v1 , = λ1 hv2 , v1 i,
hAv2 , v1 i = hλ2 v2 , v1 i = (λ2 v2 )T v1 = λ2 v2T v1 = λ2 hv2 , v1 i.
From (a) and (b) we deduce that
0 = (λ1 − λ2 )hv2 , v1 i = (λ1 − λ2 )hv2 , v1 i,
which proves that v1 and v2 are orthogonal since λ1 6= λ2 .
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Final Exam, December 13, 2011
Question 11: The following λ1 := −2,
 v1 :=
1 −3
4, v3 := (1, 1, 2) are eigenpairs of A = 3 −5
6 −6
du(t)
the ODE system dt = Au(t)?
(1,
1, 0), λ2 := −2, v2 := (1, 0, −1), and λ3 :=
3
3. (a) What are the fundamental solutions of
4
The fundamental solutions have been defined in class and are
e−2t v1 ,
e−2t v2 ,
and e4t v3 .
This is all I wanted you to write.
Recall that upon defining the matrix P := [v1T v2T v3T ], we have A = P DP −1 where D is the diagonal
matrix composed of the eigenvalues. Then the ODE can be re-written as follows:
d(P −1 u)
= DP −1 u.
dt
−1
This can also be re-written dw
u(t). Then w(t) is given by
dt = Dw by defining w(t) := P

 
 

 
α1 e−2t
0
0
1
−2t
4t
−2t
1 + α3 e 1 ,
0 + α2 e
w(t) = α2 e−2t  = α1 e
0
0
0
α3 e4t
where α1 , α2 , α3 are any real numbers. As a result,
 
 
0
1
u(t) = P w(t) = α1 e−2t P 0 + α2 e−2t P 1 + α3 e4t P
0
0
 
0
1 = α1 e−2t v1 + α1 e−2t v2 + α1 e4t v3 .
0
In conclusion, u(t) is a linear combination of the fundamental solutions e−2t v1 , e−2t v2 , e4t v3 .
(b) What is the solution of
du(t)
dt
= Au(t), with u(0) = (2, 1, −1)?
We know that u(t) is a linear combination of the fundamental solutions:
u(t) = α1 e−2t v1 + α2 e−2t v2 + α3 e4t v3 .
The condition u(0) = (2, 1, −1) implies
(2, 1, −1) = α2 v1 + α2 v2 + α3 v3 .
The corresponding

1 1
1 0
0 −1
augmented matrix is
 
 
1 2
1 1 1 2
1
1 1  ∼ 0 −1 0 −1 ∼ 0
2 −1
0 −1 2 −1
0
1
1
0
1
0
2
 
2
1
1 ∼ 0
0
0
The solution is (α1 , α2 , α3 ) = (1, 1, 0). This means
u(t) = e−2t v1 + e−2t v2 = e−2t (2, 1, −1).
0
1
0
0
0
1

1
1 .
0