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Math 241 Makeup Final Answers: Solutions to Complex Analysis Problems, Exams of Mathematics

Bhupendra Narayan Mandal UniversityMathematics

The answers to the short answer and true/false questions from a complex analysis makeup final exam. It includes calculations for finding residues, laurent expansions, and fourier series. Useful for students studying complex analysis and preparing for exams.

Typology: Exams

2012/2013

Uploaded on 02/12/2013

sathyanarayana
sathyanarayana 🇮🇳

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Math 241 Makeup Final Answers
1) True False
1. True.
2. False.
3. True.
4. False.
5. False.
6. False.
7. False.
8. True.
2) Short Answer
1. 12.
2. 2πi.
3. eiπ
4, ei3π
4, ei5π
4, ei7π
4.
4. Essential.
5. 0. The function is analytic.
6. 7. All the Fourier coefficients are zero except for 7, the coefficient on
cos 2x.
7. √2.
8. 3.
9. −4.
pf3
pf4
pf5

Partial preview of the text

Download Math 241 Makeup Final Answers: Solutions to Complex Analysis Problems and more Exams Mathematics in PDF only on Docsity!

Math 241 Makeup Final Answers

  1. True False
  1. True.
  2. False.
  3. True.
  4. False.
  5. False.
  6. False.
  7. False.
  8. True.
  1. Short Answer
  2. 2πi.
  3. ei^ π^4 , ei^34 π^ , ei^54 π^ , ei^74 π^.
  4. Essential.
    1. The function is analytic.
    1. All the Fourier coefficients are zero except for 7, the coefficient on cos 2x.
  7. √ 2.
  9. − 4.
  1. The integrand has simple poles at z = i, −i. Both poles are inside the contour C. Res(z

z^2 + 1 , i) =^

i^2 − 4 2 i =^

2 i =

2 i. Res(z

z^2 + 1 ,^ −i) =

(−i)^2 − 4 − 2 i =^

− 2 i =^ −

2 i. So Answer: (^) ∫ C

z^2 − 4 z^2 + 1 dz^ = 2πi(

2 i^ −^

2 i) = 0

  1. The integrand has a pole of order 2 at 2. The pole is inside the contour C. Res( e

z (z − 2)^2 ,^ 2) = lim^ z→^2

d dz [(z^ −^ 2)

2 ez (z − 2)^2 ] = lim^ z→^2 e

z (^) = e (^2).

So Answer: (^) ∫ C

ez (z − 2)^2 dz^ = 2πi(e

(^2) ) = 2e (^2) πi.

  1. The integrand, which we think of as (^) zsin cos^ z z , has simple poles at 0 and all odd multiples of π 2. The poles inside C are π 2 , 32 π , and 52 π. Note that z = 0 is a removable singularity and hence its residue is 0. Res( zsin cos^ z z , π 2 ) = (^0) −^1 (π 2 )

= − (^) π^2

Res( zsin cos^ z z , 32 π) = (^0) − (− 3 π^1 2 )(−1) = − (^32) π Similarly Res( (^) zsin cos^ z z , 32 π ) = − (^52) π. So Answer: ∫ C

tan z z dz^ =^ −^2 πi(

π +^

3 π +^

5 π ) =^ −^2 πi

15 π =^ −

15 i.

    1. Make the substitution z = eix^ so dx = (^) iz^1 dz and cos x = 12 (z + (^1) x ). Thus if C is the unit circle our integral becomes ∫ C

2 − 12 (z + (^1) z )

iz dz^ = 2i

C

z^2 − 4 z + 1 dz. The integrand has simple poles at z = 2 ± √ 3 , but only 2 − √3 is inside the contour ∫ C. Thus

C

z^2 − 4 z + 1 dz^ = 2πi(Res(^

z^2 − 4 z + 1,^2 −

√3) = 2πi 1 2(2 − √3) − 4 =^ −^ √πi

Thus Answer: 2 i(− √πi 3 ) = √^2 π 3.

  1. For this type of integral we compute ∫^ −∞∞ (x 2 x+1)(^2 e(iπxx 2 )+2) dx and take the real part. So

Rlim→∞

∫ R

−R

x^2 e(iπx) (x^2 + 1)(x^2 + 2) dx^ = lim R→∞

IR

z^2 e(iπz) (z^2 + 1)(z^2 + 2) dz = lim R→∞

CR

z^2 e(iπz) (z^2 + 1)(z^2 + 2) dz^ +

SR

z^2 e(iπz) (z^2 + 1)(z^2 + 2) dz

Where IR = {−R ≤ x ≤ R}, SR is the circle of radius R about the origin and CR is their union. The only singularities inside CR are two poles at i and √ 2 i. so ∫ CR

z^2 e(iπz) (z^2 + 1)(z^2 + 2) dz^ = 2πi(Res(^

z^2 e(iπz) (z^2 + 1)(z^2 + 2), i) + Res(^

z^2 e(iπz) (z^2 + 1)(z^2 + 2),^

√ 2 i))

= 2πi(−e −π 2 i +^

− 2 e−π√^2 − 2 √ 2 i ) =^ −

π(2e−π√ √^2 + √ 2 e−π) 2 Also

Rlim→∞

SR

z^2 e(iπz) (z^2 + 1)(z^2 + 2) dz^ = 0 by a theorem from class. Thus Answer: −π(2e

−π√ (^2) + √ 2 e−π) √ 2

  1. Take the Laplace transform of both sides to get L{y′′^ − 5 y′^ + 4y = 0} = 0 so s^2 ˆy − sy(0) − y′(0) − 5 syˆ + 5y(0) + 4ˆy = 0 (s^2 − 5 s + 4)ˆy = 2 yˆ = (^) s (^2) − 52 s + 4 =^23 s −^1 4 − (^23) s −^1 . Thus y(t) = L−^1 {ˆy} =^23 L−^1 { (^) s −^1 4 } − 23 L−^1 { (^) s −^1 1 }. So Answer: y(t) =^23 e^4 t^ − 23 et.
    1. Let f (z) = u(x, y) + iv(x, y) be an analytic function. The real part of f is u(x, y). uxx + uyy = (ux)x + (uy)y = (vy)x + (−vx)y = vxy − vxy = 0, where the second equality follows from the Cauchy Riemann equations.
  1. The function u(x, y) = xy −1 is clearly harmonic. If v(x, y) is its harmonic conjugate then vy = ux = y and vx = −uy = −x. Integrating the first equation with respect to y gives

v(x, y) =^12 y^2 + g(x).

Thus vx = g′(x) but this must be −x so g(x) = −^12 x^2 + c. So Answer: v(x, y) =^12 (y^2 − x^2 + c)

  1. The equation becomes X′′T − 2 XT ′′^ + 7XT ′^ + XY = 0. Dividing by XT and rearranging we get X′′ X =

T (2T^

′′ − 7 T ′) − 1.

Answer: Eigenvalues λn = (^ nπ 3 )^2 with eigenfunctions cos nπ 3 x, where n runs through all non-negative integers.

  1. We are looking for a power series solution∑∞ y(x) = ∑∞ n=0 anxn. So y′(x) = n=1 nanxn−^1 =^

n=0(n^ + 1)an+1xn^ and^ y′′(x) =^

∑∞^ n=2^ n(n^ −^ 1)anxn−^2 = n=0(n^ + 2)(n^ + 1)an+2xn.^ Thus the equation is y′′^ + 2y′^ + y =

∑^ ∞

n=

[(n + 2)(n + 1)an+2 + 2(n + 1)an+1 + an]xn^ = 0.

So Answer: an+2 = −a (nn^ −+ 2)(^ (n^ + 1)n + 1)an+.