# math of complex variable

*label*Mathematics

*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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## Tutor Answer

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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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