King Fahd University of Petroleum & Minerals Department of ... · King Fahd University of Petroleum & Minerals Department of Mathematics and Statistics MATH 302 FINAL EXAM (172) Sunday,

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.

2. Show all your work. No points for answers without justification.

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 − 𝑎0 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 = (10

) 𝑎𝑛𝑑 𝐾2 = (11

).

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

𝑃 = (1 10 1

)

and the diagonal matrix D is

𝐷 = 𝐏−1𝐀𝐏 = (𝑎 00 𝑏

).

(d) We have

𝐴𝑘 = 𝑃𝐷𝑘𝑃−1 = (1 10 1

) (𝑎 00 𝑏

)𝑘

(1 −10 1

)

= (1 10 1

) (𝑎𝑘 00 𝑏𝑘) (

1 −10 1

) = (𝑎𝑘 𝑏𝑘 − 𝑎𝑘

0 𝑏𝑘 ).

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 ∙ 𝒅𝐒𝑆

of the vector field

𝐅 = (3𝑥 + 𝑧77) 𝒂𝒙 + (𝑦2 − 𝑠𝑖𝑛 (𝑥2𝑧)) 𝒂𝒚 + (𝑥𝑧 + 𝑦𝑒𝑥5) 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 ∙ 𝒅𝑺𝑆

= ∭ div 𝐹 𝑑𝑉𝐷

= ∫ ∫ ∫ (3 + 2𝑦 + 𝑥)𝑑𝑥𝑑𝑦𝑑𝑧1

0

3

0

2

0

= ∫ ∫ (3 + 2𝑦 +1

2) 𝑑𝑦𝑑𝑧

3

0

2

0

= ∫ (9 + 9 +3

2) 𝑑𝑧

2

0

= 36 + 3 = 39

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 ∙ 𝒅𝐥𝐿

of the vector field 𝐅 = 2𝑦2𝒂𝒙 + 𝑥2𝑦 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 ∙ 𝒅𝑺𝑹𝑳

= ∫ ∫ (𝟐𝒙𝒚 − 𝟒𝒚)𝒅𝒚𝒅𝒙𝟐−𝒙

𝟎

𝟐

𝟎

=𝟏

𝟐∫ (𝟐𝒙 − 𝟒)[𝒚𝟐]𝒚=𝟎

𝟐−𝒙𝒅𝒙𝟐

𝟎

= ∫ (𝒙 − 𝟐)𝟑𝒅𝒙𝟐

𝟎

=𝟏

𝟒[(𝒙 − 𝟐)𝟒]𝒙=𝟎

𝟐

= −𝟒.

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, … .

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) 𝑑𝑧𝐶

= ∫ (2cos 𝑡 + 1) 2𝑖𝑒𝑖𝑡𝑑𝑡 𝜋

𝜋2

= 2𝑖 ∫ (2 (𝑒𝑖𝑡 + 𝑒−𝑖𝑡

2) + 1) 𝑒𝑖𝑡𝑑𝑡

𝜋

𝜋2

= 2𝑖 ∫ (𝑒2𝑖𝑡 + 1 + 𝑒𝑖𝑡) 𝑑𝑡𝜋

𝜋2

= [𝑒2𝑖𝑡 + 2𝑖 𝑡 + 2𝑒𝑖𝑡]𝜋2

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.

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= −

𝜕𝑣

𝜕𝑥,

hold only at point (0,0), then 𝑓(𝑧) is not differentiable at any 𝑧 ≠ 0.

Moreover, since 𝑢, 𝑣,𝜕𝑢

𝜕𝑥,

𝜕𝑢

𝜕𝑦,

𝜕𝑣

𝜕𝑥,

𝜕𝑣

𝜕𝑦 are continuous in a neighborhood about a point

(0,0), then 𝑓(𝑧) is differentiable at 𝑧 = 0 and 𝑓′(0) =𝜕𝑢

𝜕𝑥(0,0) + 𝑖

𝜕𝑣

𝜕𝑥(0,0) = 0.

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 | 8

Q7. Use Cauchy’s integral formulas (for higher derivatives) to evaluate

∮𝑧 − 2

𝑧2(𝑧 − 1) 𝑑𝑧

𝐶

,

where C is the contour shown in the below figure.

Solution: Observe that the points 𝑧 = 0 and 𝑧 = 1 are singularities lying within the non-simple contour C.

1 2 1 2

2 2 2 2 2

2 2 2 2 2

1 1 1 1 1C C C C C

z z z z zdz dz dz dz dz

z z z z z z z z z z

where 𝐶1 and 𝐶2 are the simple contours shown in the below figure.

Let 𝑓1(𝑧) =𝑧−2

𝑧−1. We have 𝑓1

′(𝑧) =1

(𝑧−1)2 , 𝑓1′(0) = 1. Using the Cauchy's integral formula, we obtain

∮𝑧 − 2

𝑧2(𝑧 − 1) 𝑑𝑧 = ∮

𝑓1(𝑧)

𝑧2 𝑑𝑧

𝐶1𝐶1

= 2𝜋𝑖𝑓1′(0) = 2𝜋𝑖.

Let 𝑓2(𝑧) =𝑧−2

𝑧2 We have 𝑓2(1) = −1. Using the Cauchy's integral formula, we obtain

∮𝑧 − 2

𝑧2(𝑧 − 1) 𝑑𝑧 = ∮

𝑓2(𝑧)

𝑧 − 1 𝑑𝑧

𝐶2𝐶2

= 2𝜋𝑖𝑓2(1) = −2𝜋𝑖

Hence,

∮𝑧−2

𝑧2(𝑧−1) 𝑑𝑧

𝐶= −(2𝜋𝑖) + (−2𝜋𝑖) = −4𝜋𝑖.

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 | 9

Q8. Let 𝑓(𝑧) =1

1+𝑧2.

(a) Find the Taylor series expansion of the function on the open disk |𝑧 − 1| < √2.

(b) Find the Laurent series expansion of the function on the region |𝑧 − 1| > √2.

Solution:

We want all powers of (𝑧 − 1) in the series.

𝑓(𝑧) =1

2𝑖(

1

𝑧 − 𝑖−

1

𝑧 + 𝑖)

=1

2𝑖(

1

(𝑧 − 1) + 1 − 𝑖−

1

(𝑧 − 1) + 1 + 𝑖)

Now we need to consider the two cases:

(a) |𝑧 − 1| < √2:

𝑓(𝑧) =1

2𝑖((

1

1 − 𝑖)

1

1 − (−𝑧 − 11 − 𝑖

)− (

1

1 + 𝑖)

1

1 − (−𝑧 − 11 + 𝑖

))

=1

2𝑖((

1

1 − 𝑖) ∑ (−

𝑧 − 1

1 − 𝑖)

𝑛∞

𝑛=0

− (1

1 + 𝑖) ∑ (−

𝑧 − 1

1 + 𝑖)

𝑛∞

𝑛=0

)

= ∑(−1)𝑛 ((1 − 𝑖)−𝑛−1 − (1 + 𝑖)−𝑛−1

2𝑖) (𝑧 − 1)𝑛

∞

𝑛=0

(b) |𝑧 − 1| > √2:

𝑓(𝑧) =1

2𝑖((

1

𝑧 − 1)

1

1 − (−1 − 𝑖𝑧 − 1

)− (

1

𝑧 − 1)

1

1 − (−1 + 𝑖𝑧 − 1

))

=1

2𝑖((

1

𝑧 − 1) ∑ (−

1 − 𝑖

𝑧 − 1)

𝑛∞

𝑛=0

− (1

𝑧 − 1) ∑ (−

1 + 𝑖

𝑧 − 1)

𝑛∞

𝑛=0

)

= ∑(−1)𝑛 ((1 − 𝑖)𝑛 − (1 + 𝑖)𝑛

2𝑖)

∞

𝑛=0

(𝑧 − 1)−𝑛−1

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 | 10

Q9. Let 𝑓(𝑧) =𝑠𝑖𝑛 𝑧

z2 𝑠𝑖𝑛ℎ 𝑧.

(a) Find all the poles of 𝑓(𝑧) and their orders.

(b) Use the residue theorem to evaluate ∮ 𝑓(𝑧)𝐶

𝑑𝑧 where 𝐶 is the positively oriented

circle |𝑧| = 4.

Solution:

(a) 𝑓(𝑧) =𝑠𝑖𝑛 𝑧

z2 𝑠𝑖𝑛ℎ 𝑧 has the singularities 𝑧 = 0 and 𝑧 = 𝑛𝜋𝑖 where 𝑛 = ±1, ±2, ±3, … .

Since 𝑧 = 0 is a zero of the denominator of order 3 and zero of the numerator of order 1, then 𝑧 = 0 is a double pole.

Similarly, since 𝑧 = 𝑛𝜋𝑖 is a zero of the denominator of order 1 and is not a zero of the numerator, then 𝑧 = 𝑛𝜋𝑖 is a simple pole where 𝑛 = ±1, ±2, ±3, … .

(b) The only singularities lie within C are 𝑧 = 0 and 𝑧 = ±𝜋𝑖. Thus,

𝐼 = ∮𝑠𝑖𝑛 𝑧

z2 𝑠𝑖𝑛ℎ 𝑧

𝐶

𝑑𝑧 = 2𝜋𝑖 [𝑅𝑒𝑠(𝑓(𝑧), 0) + 𝑅𝑒𝑠(𝑓(𝑧), 𝜋𝑖) + 𝑅𝑒𝑠(𝑓(𝑧), −𝜋𝑖)]

Since 𝑧 = 0 is a double pole, then

𝑅𝑒𝑠(𝑓(𝑧), 0) = 𝑙𝑖𝑚𝑧→0

𝑑

𝑑𝑧(𝑧2𝑓(𝑧)) = 𝑙𝑖𝑚

𝑧→0

𝑑

𝑑𝑧(

𝑠𝑖𝑛 𝑧

𝑠𝑖𝑛ℎ 𝑧) = 𝑙𝑖𝑚

𝑧→0

𝑠𝑖𝑛ℎ 𝑧 𝑐𝑜𝑠 𝑧 − 𝑠𝑖𝑛 𝑧 cosh 𝑧

𝑠𝑖𝑛ℎ2 𝑧

= 𝑙𝑖𝑚𝑧→0

−sin 𝑧

𝑐𝑜𝑠ℎ 𝑧= 0.

Since 𝑧 = ±𝜋𝑖 is simple pole, then

𝑅𝑒𝑠(𝑓(𝑧), 𝜋𝑖 )= 𝑙𝑖𝑚𝑧→ 𝜋𝑖

(𝑧 − 𝜋𝑖)𝑓(𝑧) = 𝑙𝑖𝑚𝑧→ 𝜋𝑖

(𝑧− 𝜋𝑖) 𝑠𝑖𝑛 𝑧

z2 𝑠𝑖𝑛ℎ 𝑧

= 𝑙𝑖𝑚𝑧→ 𝜋𝑖

(𝑧 − 𝜋𝑖) 𝑐𝑜𝑠 𝑧 + 𝑠𝑖𝑛𝑧

z2 𝑐𝑜𝑠ℎ 𝑧 + 2𝑧 𝑠𝑖𝑛ℎ 𝑧= −

𝑠𝑖𝑛( 𝜋𝑖)

𝜋2 𝑐𝑜𝑠ℎ( 𝜋𝑖)

𝑅𝑒𝑠(𝑓(𝑧), −𝜋𝑖 )= 𝑙𝑖𝑚𝑧→ −𝜋𝑖

(𝑧 + 𝜋𝑖)𝑓(𝑧) = 𝑙𝑖𝑚𝑧→ −𝜋𝑖

(𝑧+ 𝜋𝑖) 𝑠𝑖𝑛 𝑧

z2 𝑠𝑖𝑛ℎ 𝑧

= 𝑙𝑖𝑚𝑧→ −𝜋𝑖

(𝑧 + 𝜋𝑖) 𝑐𝑜𝑠 𝑧 + 𝑠𝑖𝑛𝑧

z2 𝑐𝑜𝑠ℎ 𝑧 + 2𝑧 𝑠𝑖𝑛ℎ 𝑧=

𝑠𝑖𝑛(𝜋𝑖)

𝜋2 𝑐𝑜𝑠ℎ(𝜋𝑖)

Hence, 𝐼 = 0

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 | 11

Q10. Evaluate the Cauchy principal value of ∫𝑥2 +1

𝑥4 +1

∞

−∞𝑑𝑥.

Solution:

Let 𝑓(𝑧) =𝑧2 +1

𝑧4 +1 .

Since 𝑧4 +1 = (𝑧 − 𝑧0)(𝑧 − 𝑧1)(𝑧 − 𝑧2)(𝑧 − 𝑧3) where 𝑧𝑘 = 𝑒𝜋𝑖+2𝑘𝜋𝑖

4 , 𝑘 = 0,1,2,3.

We let 𝐶 be the closed contour consisting of the interval [−𝑅, 𝑅] on the x-axis and the semi-

circle 𝐶𝑅 of radius 𝑅 > 1 given by 𝑧 = 𝑅𝑒𝑖𝑡 , 𝑡 = 0 … 𝜋.

Then

∮ 𝑓(𝑧)𝐶

𝑑𝑧 = ∫𝑥2 +1

𝑥4 +1

𝑅

−𝑅

𝑑𝑥 + ∫𝑧2 +1

𝑧4 +1 𝑑𝑧,

𝐶𝑅

Since degree 𝑧4 +1 minus degree 𝑧2 +1 ≥ 2, then 𝑙𝑖𝑚𝑅→∞

∫𝑧2 +1

𝑧4 +1 𝑑𝑧 = 0

𝐶𝑅 and so

𝑃. 𝑉. ∫𝑥2 +1

𝑥4 +1

∞

−∞𝑑𝑥=2𝜋𝑖 [𝑅𝑒𝑠(𝑓(𝑧), 𝑧0) + 𝑅𝑒𝑠(𝑓(𝑧), 𝑧1)].

Since 𝑧 = 𝑧0 and 𝑧 = 𝑧1 are simple poles, then

𝑅𝑒𝑠(𝑓(𝑧), 𝑧0) =𝑧0

2 +1

4𝑧03

=1 + 𝑖

4𝑒−

3𝜋𝑖4 = −

1

4√2(1 + 𝑖)2 = −

𝑖

2√2

𝑅𝑒𝑠(𝑓(𝑧), 𝑧1) =𝑧1

2 +1

4𝑧13

=1 − 𝑖

4𝑒−

9𝜋𝑖4 =

1 − 𝑖

4𝑒−

𝜋𝑖4 = −

𝑖

2√2

Hence,

𝑃. 𝑉. ∫𝑥2 +1

𝑥4 +16

∞

−∞𝑑𝑥 = √2 𝜋.