math of complex variable

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timer Asked: Oct 16th, 2018
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Question Description

QUIZ about harmonic conjugate and some evaluation. please do all the question and step by step.

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MTH 365 Quiz 7 Due: 2018-10-17 Name: Section: 21 Instructions: 1. Read the directions carefully. 2. Write neatly in pencil and show all your work. 3. Use the appropriate notation. 4. Do not use decimals on any intermediate step. 5. If you have trouble during the quiz, feel free to ask me for help. Score: 1. Evaluate the following. Z a. (2z + z)dz, where C is given by x = t2 , y = t, 0 ≤ t ≤ 2. C Z b. C z2 + 1 dz, where C is the right half of the circle |z| = 1 from z = −i to z = i. z 2. Evaluate the following. I a. Re(z)dz, where C is given by |z| = 1. C Z b. C (x2 + iy)dz, where C is the straight line from z = i to z = 1. Z 3. Evaluate C (z 2 + 1)dz, where C is the polygon path consisting of the line segments from z = 0 to z = 1 + i and from z = 1 + i to z = 2 + i. I (2z − 1)dz, where C is given below. 4. Evaluate C 5. Evaluate the following where C is given by |z| = 1. I a. sin(z)dz C Z  b. z2 + C 1 z−4  dz ...
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Tutor Answer

Borys_S
School: UIUC

The solutions are ready, please read them and ask if something is unclear.

1.
a. 𝐶 is 𝑥(𝑡) = 𝑡 2 , 𝑦(𝑡) = 𝑡, 0 ≤ 𝑡 ≤ 2. Then 𝑧 = 𝑥 + 𝑖𝑦, 𝑧̅ = 𝑥 − 𝑖𝑦, 𝑧 ′ (𝑡) = 2𝑡 + 𝑖 and
2

2

∫ (2𝑧̅ + 𝑧)𝑑𝑧 = ∫ (2𝑧̅(𝑡) + 𝑧(𝑡))𝑧
𝐶

2

0

2

′ (𝑡)𝑑𝑡

= ∫ (2𝑥(𝑡) − 2𝑖𝑦(𝑡) + 𝑥(𝑡) + 𝑖𝑦(𝑡))(2𝑡 + 𝑖)𝑑𝑡 =
0

2

= ∫ (3𝑡 2 − 𝑖𝑡)(2𝑡 + 𝑖)𝑑𝑡 = ∫ (6𝑡 3 − 2𝑖𝑡 2 + 3𝑖𝑡 2 + 𝑡)𝑑𝑡 = ∫ (6𝑡 3 + 𝑖𝑡 2 + 𝑡)𝑑𝑡 =
0

0

0

3 4 1 3 1 2 2
8
𝟖
= ( 𝑡 + 𝑖𝑡 + 𝑡 )
= 24 + 𝑖 + 2 = 𝟐𝟔 + 𝒊.
2
3
2
3
𝟑
𝑡=0

b. 𝐶 is 𝑧(𝑡) = −𝑖𝑒 𝑖𝑡 = sin 𝑡 − 𝑖 cos 𝑡 , 0 ≤ 𝑡 ≤ 𝜋. We see

1
𝑧(𝑡)

= 𝑖𝑒 −𝑖𝑡 = sin 𝑡 + 𝑖 cos 𝑡 and

𝑧 ′ (𝑡) = 𝑒 𝑖𝑡 , so
𝜋
𝜋
1
∫ (𝑧 + ) 𝑑𝑧 = ∫ (−𝑖𝑒 𝑖𝑡 + 𝑖𝑒 −𝑖𝑡 )𝑒 𝑖𝑡 𝑑𝑡 = 𝑖 ∫ (1 − 𝑒 2𝑖𝑡 )𝑑𝑡 =
𝑧
𝐶
0
0
2𝑖𝑡 𝜋
𝑒
1
1
= 𝑖 (𝑡 −
)
= 𝑖 (𝜋 − − 0 + ) = 𝒊𝝅.
2𝑖 𝑡=0
2𝑖
2𝑖

2.
a. The parametrization is 𝑧(𝑡) = 𝑒 𝑖𝑡 , 0 ≤ 𝑡 ≤ 2𝜋, 𝑅𝑒(𝑧(𝑡)) = cos 𝑡 , 𝑧 ′ (𝑡) = 𝑖𝑒 𝑖𝑡 and
2𝜋

2𝜋

2𝜋

∫ 𝑅𝑒(𝑧(𝑡))𝑑𝑧 = ∫ cos 𝑡 ∙ 𝑖(cos 𝑡 + 𝑖 sin 𝑡)𝑑𝑡 = 𝑖 ∫ cos2 𝑡 𝑑𝑡 − ∫ sin 𝑡 cos 𝑡 𝑑𝑡 =
𝐶

0

0

0

𝑖 2𝜋
1 2𝜋
𝑖
= ∫ (1 + cos(2𝑡))𝑑𝑡 − ∫ sin(2𝑡) 𝑑𝑡 = ∙ (2𝜋 + 0) − 0 = 𝒊𝝅.
2 0
2 0
2

b. The parametrization is 𝑧(𝑡) = 𝑖(1 − 𝑡) + 𝑡, 0 ≤ 𝑡 ≤ 1. This way 𝑥(𝑡) = 𝑡, 𝑦(𝑡) = 1 − 𝑡,
𝑧 ′ (𝑡) = 1 − 𝑖 and 𝑥 2 (𝑡) + 𝑖𝑦(𝑡) = 𝑡 2 + 𝑖(1 − 𝑡), so the integral is
1
1
1
𝑖
∫ (𝑡 2 + 𝑖(1 − 𝑡))(1 − 𝑖)𝑑𝑡 = (1 − 𝑖) ( 𝑡 3 − 𝑡 2 + 𝑖𝑡)
=
3
2
𝑡=0
0
1 𝑖
1 1
1 1
1
1 𝟓 𝟏
= (1 − 𝑖) ( − + 𝑖) = (1 − 𝑖) ( + 𝑖) = − 𝑖 + 𝑖 + = + 𝒊.
3 2
3 2
3 3
2
2 𝟔 𝟔

3. This integral consists of two parts, integral over the first segment and over the second
segment. [we can add the third segment from 2 + 𝑖 to 0 instead and use Cauchy-Goursat
theorem. But it seems simpler to do two integrals.]
3.1. 𝑧(𝑡) = (1 + 𝑖)𝑡, 0 ≤ 𝑡 ≤ 1. Then 𝑧 ′ (𝑡) = 1 + 𝑖 and the integral is
1

1

2

∫ (((1 + 𝑖)𝑡) + 1) (1 + 𝑖)𝑑𝑡 = (1 + 𝑖) ∫ ((1 + 𝑖)2 𝑡 2 + 1)𝑑𝑡 =
0

1

0

(1 + 𝑖)2 3
(1 + 𝑖)2
𝟏
= (1 + 𝑖) (
𝑡 + 𝑡)
= (1 + 𝑖) (
+ 1) = (𝟏 + 𝒊)𝟑 + 𝒊 + 𝟏.
3
3
𝟑
𝑡=0

3.2. Here 𝑧(𝑡) = (1 − 𝑡)(1 + 𝑖) + 𝑡(2 + 𝑖) = 1 + 𝑖 + 𝑡, 0 ≤ 𝑡 ≤ 1. Then 𝑧 ′ (𝑡) = 1 and the
integral is
1
1
1
∫ ((1 + 𝑖 + 𝑡)2 + 1)𝑑𝑡 = ( (1 + 𝑖 + 𝑡)3 + 𝑡)
=
3
𝑡=0
0
1
1
𝟏
𝟏
= (2 + 𝑖)3 + 1 − (1 + 𝑖)3 − 0 = (𝟐 + 𝒊)𝟑 + 𝟏 − (𝟏 + 𝒊)𝟑 .
3
3
𝟑
𝟑
3.3. (adding). So, the entire integral is equal to
1
1
1
1
(1 + 𝑖)3 + 𝑖 + 1 + (2 + 𝑖)3 + 1 − (1 + 𝑖)3 = 𝑖 + 1 + (2 + 𝑖)3 + 1 =
3
3
3
3
1
𝟖 𝟏𝟏
= 2 + 𝑖 + (8 + 12𝑖 − 6 − 𝑖) = +
𝒊.
3
𝟑 𝟑

4. The integral is zero by Cauchy-Goursat theorem (the function is analytic everywhere
and the contour is suitable, too).

5.
a. The integral is zero by Cauchy-Goursat theorem, too.

b. This integral is zero, too, because the function IS analytic inside on a realm containing
the given contour and its interior, say, inside |𝑧| = 2.
[it could be nonzero if the contour would contain the point ...

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Anonymous
awesome work thanks

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