Last name:
name:
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−x (x sin(y) − y cos(y)) + ie−x (y sin(y) − x cos(y)),
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−x (x sin(y) − y cos(y)) and Q(x, y) =
e−x (y sin(y) − x cos(y))
∂x P (x, y) = e−x (−x sin(y) + y cos(y) + sin(y))
∂y Q(x, y) = e−x (sin(y) + y cos(y) + x sin(y))
Clearly ∂x P (x, y) 6= ∂y Q(x, y); as a consequence, f is not holomorphic.
R
Question 2: Evaluate C zdz along the curve C = γ([0, 2]) where γ(t) = t + it2 .
Since dz = γ 0 (t)dt = (1 + 2it)dt, we have
2
Z
Z
2
(t + it )(1 + 2it)dt =
zdz =
C
Z
0
=−
32
+ 8i + 2 = −6 + 8i.
4
0
2
(−2t3 + 3it2 + t)dt
2
Final Exam, December 15, 2010
Question 3: (a) Is the function f (z) = z holomorphic? Justify your answer.
From class we know that f is not holomorphic. One way to see this is to verify the Cauchy-Riemann
relations do not hold. Let x = P (x, y) = Re(f (z)) and −y = Q(x, y) = Im(f (z)). Then
∂x P (x, y) = 1 6= −1 = ∂y Q(x, y).
(b) Let C be the unit circle centered at 0 in C. Compute
R
zdz. (Hint: C is the unit circle).
R
On the unit circle z =
Hence the problems amounts to compute C z1 dz. The function z1 is
holomorphic in C except at 0. We can apply the Residue Theorem. The residue of z1 at 0 is 1,
obviously. In conclusion
Z
Z
1
zdz =
dz = 2iπ.
z
C
C
C
1
z.
(c) Let C be the unit circle centered at 0 in C. Compute
R
1/zdz. (Hint: C is the unit circle).
R
On the unit circle
= z. Hence the problems amounts to compute C zdz. The function z is
holomorphic in C. Cauchy’s theorem implies that the integral is zero:
Z
Z
1
dz =
zdz = 0.
C z
C
1
z
C
Last name:
name:
Question 4: Let t ∈ R, t ≥ 0. Evaluate
3
R +∞
1/|z 2 + 1| ≤ 1/(|z|2 − 1), ∀z ∈ C such that |z|
eitx
dx.
−∞ 1+x2
eitz
> 1; 1+z2 has
(Hint: |eitz | ≤ 1 if Im(z) ≥ 0;
two simple poles ±i.)
itz
e
Consider the function f (z) = 1+z
2 and consider the closed path made by the segment [−R, +R]
and the half circle of radius R centered at 0 in the top half complex plane. Let C be this closed
simple path and let CR be the top half circle of radius R centered at 0. Then
Z
+∞
−∞
eitx
= lim
R→+∞
1 + x2
and
Z
Z
+R
f (x)dx,
−R
+R
f (z)dz =
C
Z
Z
f (x)dx +
f (z)dz.
−R
CR
The integral along CR is bounded as follows:
Z
Z
Z π
|eitz |
1
R
f (z)dz ≤
|dz| ≤
Rdθ = π 2
2|
2−1
|1
+
z
|z|
R
−1
CR
CR
0
As a result
Z
lim R→+∞
CR
f (z)dz = 0.
The pole i is in the region enclosed by C. The residue of f at i is given by
Res(i) = [f (z)(z − i)](z = i) = [
eitz
e−t
](z = i) =
.
z+i
2i
Finally
Z
f (z)dz = 2iπRes(i) = 2iπ
C
e−t
= πe−t .
2i
This implies
Z
+∞
−∞
eitx
= lim
R→+∞
1 + x2
= πe−t .
Z
+R
−R
Z
Z
f (z)dz − lim
f (x)dx = lim
R→+∞
C
R→+∞
f (z)dz
CR
4
Final Exam, December 15, 2010
1 7
. Is it possible to find a set of two independent
7 −3
eigenvectors of A? Be very accurate and explicit when answering.
Question 5: Consider the matrix A =
Since the matrix is symmetric, it is diagonalizable, which implies that, yes indeed, one can find a
set of two independent eigenvectors of A.
π
0
Question 6: Find the eigenvalues of
and give two orthogonal eigenvectors.
0 log(2)
The matrix is diagonal. The eigenvalues are π and log(2). The vectors (1, 0)T and (0, 1)T are
eigenvectors of associated with π and log(2), respectively and they are obviously orthogonal.
Question 7: Let A be a n×n real-valued symmetric matrix, n ≥ 2. 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 scalar product (v, w) 7→ wT v.
We have
(v2 , Av1 ) := v2T Av1 = v2T (λ1 v1 ) = λ1 v2T v1 ,
(v1 , Av2 ) := v1T Av2 = v1T (λ2 v2 ) = λ2 v1T v2 .
But, A being symmetric means that
(v2, Av1 ) := v2T Av1 = (v2T Av1 )T = v1T AT v2 = v1T Av2 := (v1 , Av2 ).
As a result
λ1 v2T v1 = λ2 v1T v2 = λ2 v2T v1 .
This implies (λ1 − λ2 )v2T v1 = 0. This is possible only if v2T v1 = 0 since λ1 6= λ2 .
Last name:
name:
Question 8: Consider the dynamical system
du
dt
=
5
2
1
4
u. Is this dynamical system stable?
−1
The two eigenvalues of the matrix are the roots of
0 = (2 − λ)(−1 − λ) − 4 = λ2 − λ − 6.
The two roots are λ1 = 3 and λ2 = −2. The dynamical system is unstable since λ1 > 0.
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
A is 5 × 5 real-valued matrix with eigenvalues −1 + 3i, −1 − 3i, − 31 + 7i,
by dz
dt = Az where
√
1
− 3 − 7i, and − 2. Is u0 a stable equilibrium solution? Justify your answer.
The real part of the five eigenvalues are negative. As a result u0 is a stable equilibrium point.
1 2
Question 10: Find the eigenvalues of
and say if the matrix is diagonalizable in R2 .
−2 1
The characteristic polynomial is
(1 − λ)2 + 4 = 0.
The two eigenvalues are complex conjugate λ1 = 1+2i, λ2 = 1−2i. The matrix is not diagonalizable
in R2 . It is diagonalizable in C though.
Question 11: Bob has written a computer code to compute eigenvalues of matrices. He tests
it on a 3×3 real-valued and obtains 2, −1 + 2i, −1 − 3i. Can anything be deduced from this
test? Justify your answer.
The computer code of Bob is wrong since the eigenvalues of a real-values matrix are complex
conjugate if they are not real. The eigenvalues are the roots of the characteristic polynomial whose
coefficients are real.
6
Final Exam, December 15, 2010
Question 12: Consider the following basis of P3 (t), S := {1, t, t2 , t3 }, and the following basis of
R3 , S 0 := {u1 , u2 , u3 } := {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Let F : P3 −→ R3 be the linear mapping
R +1
R +1
R +1
defined by F (p) = ( −1 p(t)dt, −1 p(t)tdt, −1 p(t)t2 dt).
(a) Find the matrix of F in the bases S and S 0 .
We compute the coordinate vectors of F (1), F (t), F (t2 ) and F (t3 ) relative to S 0 .
+1
Z
F (1) = (
tdt,
−1
+1
F (t) = (
−1
t2 dt,
−1
t3 dt,
−1
+1
t3 dt,
−1
This means that
Z
2
t3 dt) = (0, , 0)
3
+1
Z
−1
+1
t4 dt,
−1

[F ]SS 0
2
t2 dt) = (2, 0, )
3
+1
Z
+1
Z
−1
F (t3 ) = (
t2 dt,
−1
+1
Z
−1
+1
Z
tdt,
Z
F (t2 ) = (
+1
Z
dt,
−1
Z
+1
Z
2
= 0
2
3
0
2
3
0
Z
+1
2
2
t4 dt) = ( , 0, )
3
5
−1
2
t5 dt) = (0, , 0)
5
2
3
0
0
2
5

2
5 .
0
(b) Is the mapping F defined above bijective? Justify your answer.
The mapping F is not bijective since the dimension of the domain (dimP3 = 4) is not equal to the
dimension of the co-domain (dimR3 = 3). (Note that the rank of [F ]SS 0 is three. This means that
F is injective but is not surjective.)
Last name:
name:
7
Question 13: Let A be a m×n complex-valued matrix, m > n. Let F : Cn −→ Cm defined as
follows: F (w) = Aw. (a) Can F be surjective?
Can F cannot be surjective since the dimension of Cm , m, is larger than the dimension of Cn , n.
(b) Assume that the rank of A is equal to n. What is the kernel of F ?
It is clear that Ker(F ) = Null(A). Since the rank of A is n, Null(A) = {0}. This implies
Ker(F ) = {0}.