MAT-247 Midterm-1 Date:03/11/2018
Name Surname: ID Number:
Instructions: You have hours to complete the exam.
Question Score
1
2
3
4
5
6
Total
Notation:
R(z) represents the real part of z,
I(z) represents the imaginary part of z,
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MAT-247 Midterm-1: Complex Analysis and Vector Calculus Exercises, Exams of Engineering Mathematics
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Typology: Exams
2018/2019
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MAT-247 Midterm-1 Date:03/11/
Name Surname: ID Number:
Instructions: You have hours to complete the exam.
Question Score
Total
Notation:
R(z) represents the real part of z,
I(z) represents the imaginary part of z,
1)(10pts) Show that the angle between vectors a = [− 2 , 1 , x] and b = [x, − 1 , 2] is
always greater than π/2 for any x.
Solution: By definition of inner product:
|a · b| = |a||b| cos θ,
where θ is the angle between a and b. Then, since
|a · b| = − 2 x − 1 + 2x = − 1.
and |a| =
4 + x^2 + 1 and |b| =
4 + x^2 + 1
cos(θ) =
|a · b|
|a||b|
4 + x^2 + 1
Since x
2 ≥ 0, whatever x, then, cos(θ) < 0 for any x. Since cos(θ) < 0
while π/ 2 < θ ≤ π, θ > π/2 for any x.
3)(15pts) Find the work done in moving a particle in a force field F = x
2 i+2yzj+
y
2 k along the curve described as r(t) = (t
2 , t− 3 , t) for 1 ≤ t ≤ 2. (Hint:
Check whether the field is conservative or not).
Solution-1: Note
curlF =
i j k ∂ ∂x
∂ ∂y
∂ ∂z
x
2 2 yz y
2
curlF =
∂(y
2 )
∂y
∂(2yz)
∂z
i−
∂(y
2 )
∂x
∂(x
2 )
∂z
j+
∂(2yz)
∂x
∂(x
2 )
∂y
k = 0,
therefore F is a conservative field. Then, there exists a function f such
that ∂f
∂x
= x
2 ,
∂f
∂y
= 2yz
∂f
∂z
= y
2 ,
∂f
∂z
= y
2 → f (x, y, z) = h(x, y) + y
2 z.
Then,
∂f
∂y
∂h(x, y) + y
2 z
∂y
= g(x) + 2yz
∂f
∂x
∂g(x)
∂x
= x
2 ,
f (x, y, z) =
x
3
- y
2 z.
Then, ∫
F · dr = f (r(2)) − f (r(1)) =
Solution-2: (^) ∫
F · rdr =
∫ (^) t=
t=
F(r(t)) · r
′ (t)dt
∫ (^) t=
t=
(t
4 , 2 t(t−3), (t−3)
2 )·(2t, 1 , 1)dt =
∫ (^) t=
t=
(2t
5 +2t(t−3)+(t−3)
2 )dt
∫ (^) t=
t=
2 t
5 +2t(t−3)+(t−3)
2 )dt =
t
6
t
3
− 3 t
2
(t − 3)
3
4)(20) Compute ∮
C
y
2 dx + 3xydy,
where C is the CCW-oriented boundary of upper-half unit disk D.
(Hint: The region D is described as − 1 ≤ x ≤ 1 and 0 ≤ y ≤
1 − x^2.
The curve C is described as r(t) = (cos(t), sin(t)), for 0 ≤ t ≤ π,
and r(t) = (cos(t), 0), −π ≤ t ≤ 0. You can compute using either
description of region or description of curve).
D
C
Solution:
∮
C
y
2 dx + 3xydy =
D
∂F 2
∂x
∂F 1
∂y
dA
∂F 2
∂x
= 3y
∂F 1
∂y
= 2y
− 1
∫ √ 1 −x 2
0
ydydx =
− 1
y
2
√ 1 −x^2 0
− 1
1 − x
2
dx =
- Let z 1 and z 2 be two complex numbers. Then,
a)(15 pts) show that ∣ ∣ ∣ ∣
z 1 − z 2
z 1 + z 2
|z 2 |
|z 1 | + |z 2 |
(Hint: for any complex number z, |z| = |z|)
b)(5pts) show that ∣ ∣ ∣ ∣
z 1 − z 2
z 1 + z 2
if |z 1 | = 3|z 2 |.
Solution:
a) By the triangle inequality,
|z 1 | = |z 1 + z 2 − z 2 | ≤ |z 1 + z 2 | + |z 2 | → |z 1 | − |z 2 | ≤ |z 1 − z 2 |,
in addition, since
|z 1 + z 2 | ≤ |z 1 | + |z 2 |,
Then,
z 1 − z 2
z 1 + z 2
|z 1 | − |z 2 |
|z 1 | + |z 2 |
|z 2 |
|z 1 | + |z 2 |
|z 2 |
|z 1 | + |z 2 |
b) Since ∣ ∣ ∣ ∣
z 1 − z 2
z 1 + z 2
|z 1 | − |z 2 |
|z 1 | + |z 2 |
and
|z 2 | = |z 2 |, |z 1 | = |z 1 |,
then, (^) ∣ ∣ ∣ ∣
z 1 − z 2
z 1 + z 2
2 |z 2 |
4 |z 2 |