# Last name: name: 1

```Last name:
name:
1
Final Exam M601 (Notes, books, and calculators are not authorized) Show all your work in the
blank space on the exam sheet. Answers with no justification will not be graded.
Question 1: Let z = x + iy, were x, y ∈ R. Is the function f (z) = cos(y) + i sin(x) holomorphic
Let P (x, y) = cos(y) and Q(x, y) = sin(x) be the real and imaginary parts of f . The functions P
and Q are twice continuously differentiable functions with respect to x and y over R2 . Let us verify
whether the Cauchy Riemann relations hold:
∂x P (x, y) = 0,
∂y P (x, y) = − sin(y),
∂y Q(x, y) = 0.
∂y Q(x, y) = cos(x).
The function f is not holomorphic since ∂y P (x, y) 6= −∂x Q(x, y).
Question 2: Using the notation z = reiθ , were r ≥ 0, θ ∈ [−π, π), we define the function
log(z) as follows log(z) := log(r) + iθ. Prove that log(z) is not continuous across the axis
{y = 0, x &lt; 0}.
Let −r0 be an arbitrary number on the negative real axis {y = 0, x &lt; 0}. Let us set z+ (π − ) =
r0 ei(π−) , z− () = r0 ei(−π+) with &gt; 0. It is clear that
lim z+ () = −r0 = lim z− (),
→0
→0
i.e., both z+ () and z− () converge to −r0 as goes to zero. If the complex log 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(r0 ) + i(π − ) = log(r0 ) + iπ
→0
→0
and
lim log(z− ) = lim log(r0 ) + i(−π + ) = log(r0 ) − iπ.
→0
→0
In conclusion lim→0 log(z+ ) 6= lim→0 log(z− ), thereby proving that log(z) as defined above is not
continuous across the axis {y = 0, x &lt; 0}.
2
Final Exam, December 12, 2012
Question 3: Let z = x + iy, were x, y ∈ R. Let ez be defined by ez := ex (cos(y) + i sin(y)).
(a) Prove that eiz = e−iz by using the definition of ez .
By definition
eiz = eix−y = e−y (cos(x) + i sin(x)) = e−y (cos(x) − i sin(x)) = e−y−ix = e−i(x−iy) .
As a result eiz = e−iz
(b) Define sin(z) :=
eiz −e−iz
.
2i
Compute sin(iy) for y ∈ R.
Using the definition of sin(z), we infer that
sin(iy) =
Question 4: Evaluate
R
C
1 −y
(e − ey ) = −i sinh(y).
2i
3z 2 dz along the curve C = γ([0, 1]) where γ(t) = t2 .
Answer 1: Since dz = γ 0 (t)dt = 2tdt, we have
Z
3z 2 dz =
C
Z
1
3t4 2tdt =
0
1
6t5 dt = t6 |10 = 1.
0
0
2
Z
3
Answer 2: Observe that 3z = f (z) where f (z) = z ; as a result
Z
Z
3z 2 dz =
f 0 (z)dz = f (1) − f (0) = 1
C
C
R
Question 5: Evaluate C cos(z)dz 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) + π(t − 2) for
2 ≤ t ≤ 3.
Observe that cos(z) = f 0 (z) where f (z) = sin(z) and f is clearly holomorphic over C, (known from
class). As a result
Z
Z
cos(z)dz =
f 0 (z)dz = f (γ(3)) − f (γ(0)) = f (π) − f (0) = sin(π) − sin(0) = 0.
C
C
Last name:
name:
3
Question 6: Let f (z) = (2z 2 + 3z + 1)2 z −3 . Compute
2
1
3 π centered at 2 traversed counterclockwise.
R
C
f (z)dz where C is the circle of radius
f (z) = (2z 4 + 9z 2 + 1 + 12z 3 + 4z 2 + 6z)z −3 = (2z 4 + 12z 3 + 13z 2 + 6z + 1)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) = 13. The open
domain limited by the circle of radius 32 π centered at 12 contains the pole of f ; as a result
Z
f (z)dz = 2iπRes(f, 0) = 26iπ.
C
R 2π
iθ
Question 7: 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) =
1
2
2
3i
2i( 2 − 2)
z→ 2
Finally,
I = 2π
1
2π
=
3i
3
1
2
is in the unit circle:
1
2
4
Final Exam, December 12, 2012
Question 8: Is the matrix A =
2+i
7−i
0
2+i
There is a unique eigenvalue λ = 2 + i.
Answer 1: If A was diagonalizable it could be written as follows
A = P DP −1
where D = (2 + i)I. That would mean that
A = (2 + i)P IP −1 = (2 + i)P P −1 = (2 + i)I,
which is wrong. In conclusion A is not diagonalizable.


1 2 0
Question 9: 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&times;3 matrix A has three distinct eigenvalues, hence A is diagonalizable. (Theorem from class)
Question 10: 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 &times; 5 real-valued matrix with eigenvalues −2 + 4i, −2 − 4i,
3
10
− 13 − 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.
Last name:
name:
5
Question 11: Let A be a n&times;n complex-valued Hermitian matrix (i.e., AT = A), n ≥ 2. Let M
be a real-valued symmetric positive definite matrix. We define a new complex inner product
hv, wi := v T M w and the associated norm kvk2 := v T M v. Assume that AM = M A.
(a) Show that hv2 , Av1 i = hAv2 , v1 i, for all vectors v1 , v2 in Cn . Give all the details.
hv2 , Av1 i = v2T M Av1 = v2T M Av1 = v2T M T AT v1 = v2T AT M T v1 = (Av2 )T M 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 M v = λv T M v = λkvk2
and
hv, Avi = v T M Av = v T M 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 M w.
We have
hv2 , Av1 i = hv2 , λ1 v1 i = v2T M (λ1 v1 ) = λ1 v2T M v1 , = λ1 hv2 , v1 i,
hAv2 , v1 i = hλ2 v2 , v1 i = (λ2 v2 )T M v1 = λ2 v2T M 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 .
6
Final Exam, December 12, 2012
Question 12: The following λ1 := 1, v1 := (1, 1, 0), λ2 := −2, v2 := (1, 0, −1), and λ3 :=
3, v3 := (1, 2, 3) are eigenpairs of a matrix A. (a) What are the fundamental solutions of the
ODE system du(t)
dt = Au(t)?
The fundamental solutions have been defined in class and are
e t v1 ,
e−2t v2 ,
and e3t 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
= Dw by defining w(t) := P −1 u(t). Then w(t) is given by


 
 
 
α1 et
1
0
0
w(t) = α2 e−2t  = α1 et 0 + α2 e−2t 1 + α3 e3t 1 ,
0
0
0
α3 e3t
This can also be re-written
dw
dt
where α1 , α2 , α3 are any real numbers. As a result,
 
 
0
1
u(t) = P w(t) = α1 et P 0 + α2 e−2t P 1 + α3 e3t P
0
0
 
0
1 = α1 et v1 + α1 e−2t v2 + α1 e3t v3 .
0
In conclusion, u(t) is a linear combination of the fundamental solutions et v1 , e−2t v2 , e3t v3 .
(b) What is the solution of
du(t)
dt
= Au(t), with u(0) = (6, 5, 1)?
We know that u(t) is a linear combination of the fundamental solutions:
u(t) = α1 et v1 + α2 e−2t v2 + α3 e3t v3 .
The condition u(0) = (6, 5, 1) implies
(6, 5, 1) = α1 v1 + α2 v2 + α3 v3 .
The corresponding augmented

 
1 1 1 6
1 1
1 0 2 5 ∼ 0 −1
0 −1
0 −1 3 1
matrix is
 
1 6
1
1 −1 ∼ 0
3 1
0
1
1
0
1
−1
2
 
6
1
1 ∼ 0
2
0
The solution is (α1 , α2 , α3 ) = (3, 2, 1). This means
u(t) = 3et v1 + 2e−2t v2 + e3t v3 .
1
1
0
0
0
1
 
5
1
2 ∼ 0
1
0
0
1
0
0
0
1

3
2 .
1
```