 # King Fahd University of Petroleum & Minerals Department of ... · PDF file King Fahd University of Petroleum & Minerals Department of Mathematics and Statistics MATH 302 FINAL...

Mar 07, 2020

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• King Fahd University of Petroleum & Minerals

Department of Mathematics and Statistics

MATH 302

FINAL EXAM (172)

Sunday, May 6, 2018 Allowed Time: 3 Hours

Name: ___________________________________________________________

ID Number: _______________________ Serial Number: _______________

Section Number: _____________ Instructor’s Name: _________________

Instructions:

1. Write neatly and legibly. You may lose points for messy work.

3. Calculators and Mobiles are not allowed.

4. Make sure that you have 10 different problems (11 pages).

Problem No. Points Maximum Points

1 14

2 11

3 11

4 14

5 16

6 14

7 11

8 14

9 20

10 15 Total: 140

• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 2

Q1. Consider the 2×2 complex matrix

𝐴 = ( 𝑎 𝑏 − 𝑎 0 𝑏

).

(a) Find the eigenvalues of 𝐴. (b) For each eigenvalue of 𝐴, determine the eigenvectors. (c) Diagonalize the matrix 𝐴. (d) Using the result of the diagonalization, compute and simplify 𝐴𝑘 for each positive integer 𝑘.

Solution:

(a) Since 𝐴 is an upper triangular matrix, eigenvalues are diagonal entries.

Hence 𝑎, 𝑏 are eigenvalues of 𝐴.

(b) The corresponding eigenvectors are

𝐾1 = ( 1 0

) 𝑎𝑛𝑑 𝐾2 = ( 1 1

).

(c) Thus, a matrix P that diagonalizes A is given by

𝑃 = ( 1 1 0 1

)

and the diagonal matrix D is

𝐷 = 𝐏−1𝐀𝐏 = ( 𝑎 0 0 𝑏

).

(d) We have

𝐴𝑘 = 𝑃𝐷𝑘𝑃−1 = ( 1 1 0 1

) ( 𝑎 0 0 𝑏

) 𝑘

( 1 −1 0 1

)

= ( 1 1 0 1

) (𝑎 𝑘 0

0 𝑏𝑘 ) (

1 −1 0 1

) = (𝑎 𝑘 𝑏𝑘 − 𝑎𝑘

0 𝑏𝑘 ).

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• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 3

Q2. Let D be the region described by the box 0 < 𝑥 < 1, 0 < 𝑦 < 3, 0 < z < 2 .

Find the outward flux ∫ 𝐅 ∙ 𝒅𝐒 𝑆

of the vector field

𝐅 = (3𝑥 + 𝑧77) 𝒂𝒙 + (𝑦 2 − 𝑠𝑖𝑛 (𝑥2𝑧)) 𝒂𝒚 + (𝑥𝑧 + 𝑦𝑒

𝑥5) 𝒂𝒛 ,

where S is the boundary of D.

Solution:

Given the ugly nature of the vector field, it would be hard to compute this integral directly.

However, the divergence of 𝑭 is nice:

div 𝐹 = 3 + 2𝑦 + 𝑥.

By the divergence theorem,

∫ 𝑭 ∙ 𝒅𝑺 𝑆

= ∭ div 𝐹 𝑑𝑉 𝐷

= ∫ ∫ ∫ (3 + 2𝑦 + 𝑥)𝑑𝑥𝑑𝑦𝑑𝑧 1

0

3

0

2

0

= ∫ ∫ (3 + 2𝑦 + 1

2 ) 𝑑𝑦𝑑𝑧

3

0

2

0

= ∫ (9 + 9 + 3

2 ) 𝑑𝑧

2

0

= 36 + 3 = 39

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• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 4

Q3. Let D be the region shown in the below figure. Use Stokes’s theorem to find the circulation

∫ 𝐅 ∙ 𝒅𝐥 𝐿

of the vector field 𝐅 = 2𝑦2𝒂𝒙 + 𝑥 2𝑦 𝒂𝒚 around the positively oriented curve L

where L is the boundary of the region D.

Solution:

𝒅𝑺 = 𝒅𝒙𝒅𝒚 𝒂𝒛

𝒄𝒖𝒓𝒍 𝑭 = (𝟐𝒙𝒚 − 𝟒𝒚) 𝒂𝒛

By the Stokes’s theorem,

∫ 𝑭 ∙ 𝒅𝒍 = ∬ 𝒄𝒖𝒓𝒍 𝑭 ∙ 𝒅𝑺 𝑹𝑳

= ∫ ∫ (𝟐𝒙𝒚 − 𝟒𝒚)𝒅𝒚𝒅𝒙 𝟐−𝒙

𝟎

𝟐

𝟎

= 𝟏

𝟐 ∫ (𝟐𝒙 − 𝟒)[𝒚𝟐]𝒚=𝟎

𝟐−𝒙𝒅𝒙 𝟐

𝟎

= ∫ (𝒙 − 𝟐)𝟑𝒅𝒙 𝟐

𝟎

= 𝟏

𝟒 [(𝒙 − 𝟐)𝟒]𝒙=𝟎

𝟐

= −𝟒.

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• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 5

Q4. Solve the equation 2 cosh 𝑧 + 10 sinh 𝑧 = 5.

Solution:

Using

cosh 𝑧 = 𝑒𝑧 + 𝑒−𝑧

2 , sinh 𝑧 =

𝑒𝑧 − 𝑒−𝑧

2 ,

we obtain

𝑒𝑧 + 𝑒−𝑧 + 5(𝑒𝑧 − 𝑒−𝑧) = 5,

6𝑒𝑧 − 4𝑒−𝑧 − 5 = 0,

6𝑒2𝑧 − 5𝑒𝑧 − 4 = 0,

(3𝑒𝑧 − 4)(2𝑒𝑧 + 1) = 0,

Which implies

𝑒𝑧 = 4

3 𝑜𝑟 𝑒𝑧 = −

1

2 .

Therefore,

𝑧 = 𝑙𝑛 ( 4

3 ) = 𝑙𝑜𝑔𝑒 (

4

3 ) + 2𝑛𝜋𝑖,

or

𝑧 = 𝑙𝑛 (− 1

2 ) = 𝑙𝑜𝑔𝑒 (

1

2 ) + (𝜋 + 2𝑛𝜋)𝑖,

where 𝑛 = 0, ±1, ±2, ±3, … .

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• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 6

Q5. Evaluate the following integrals:

(a) ∫ 𝑅𝑒(𝑧 + 1) 𝑑𝑧, 𝐶

where 𝐶 is the circular arc (in the second quadrant) along |𝑧| = 2

from 𝑧 = 2𝑖 to 𝑧 = −2.

(b) ∮ 1

𝑧(𝑧+1−3𝑖) 𝑑𝑧

𝐶 , where 𝐶 is the positively oriented circle |𝑧 + 2| = 1.

Solution:

(a) On curve C, we have 𝑧 = 2cos 𝑡 + 2𝑖 sin 𝑡 = 2𝑒𝑖𝑡 , 𝜋/2 ≤ 𝑡 ≤ 𝜋, and 𝑑𝑧 = 2𝑖𝑒𝑖𝑡𝑑𝑡.

Thus,

∫ 𝑅𝑒(𝑧 + 1) 𝑑𝑧 𝐶

= ∫ (2cos 𝑡 + 1) 2𝑖𝑒𝑖𝑡𝑑𝑡 𝜋

𝜋 2

= 2𝑖 ∫ (2 ( 𝑒𝑖𝑡 + 𝑒−𝑖𝑡

2 ) + 1) 𝑒𝑖𝑡𝑑𝑡

𝜋

𝜋 2

= 2𝑖 ∫ (𝑒2𝑖𝑡 + 1 + 𝑒𝑖𝑡) 𝑑𝑡 𝜋

𝜋 2

= [𝑒2𝑖𝑡 + 2𝑖 𝑡 + 2𝑒𝑖𝑡]𝜋 2

𝜋

= (1 + 2𝑖 𝜋 − 2) − (−1 + 𝑖 𝜋 + 2𝑖)

= 𝑖 (𝜋 − 2).

(b) Both singularities 𝑧 = 0 and 𝑧 = −1 + 3𝑖 are located outside the contour 𝐶.

Therefore, ∮ 1

𝑧(𝑧+1−3𝑖) 𝑑𝑧

𝐶 = 0, by Cauchy-Goursat Theorem.

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• M A T H 3 0 2 - T e r m 1 7 2 F I N A L E X A M P a g e | 7

Q6. Let 𝑓(𝑧) = 𝑖|𝑧|2. Show that 𝑓(𝑧) is differentiable only at 𝑧0 = 0 and find 𝑓′(0).

Solution:

Let 𝑧 = 𝑥 + 𝑖 𝑦. We have

𝑓(𝑧) = 𝑖|𝑧|2 = 𝑖(𝑥2 + 𝑦2).

Put 𝑢 = 0 and 𝑣 = 𝑥2 + 𝑦2. We obtain

𝜕𝑢

𝜕𝑥 = 0,

𝜕𝑣

𝜕𝑥 = 2𝑥,

𝜕𝑢

𝜕𝑦 = 0,

𝜕𝑣

𝜕𝑦 = 2𝑦,

Since the Cauchy-Riemann equations

𝜕𝑢

𝜕𝑥 =

𝜕𝑣

𝜕𝑦 ,

𝜕𝑢

𝜕𝑦 = −

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