take home 1 solutions

8/3/2019 Take Home 1 Solutions

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Take Home 1 - Solutions

1. Assume

ez =n=0

zn

n!, cos z =

n=0

(1)2nz2n(2n)!

, sin z =n=0

(1)2n+1z2n+1(2n + 1)!

.

Show the identities,

eiz = cos z + i sin z, cos z =eiz + eiz

2, sin z =

eiz eiz2i

using the series expansions.

ANSWERS: Note

i2n = (1)n, i2n+1 = i(1)n, (i)2n = (1)n, (i)2n+1 = (i)(1)n.So, for example,

eiz + eiz

2=

1

2(n=0

(iz)n

n!+

n=0

(iz)nn!

) =

1

2(

n=0

(1)nz2n

(2n)!+

n=0

i(1)nz2n+1

(2n + 1)!+

n=0

(1)nz2n

(2n)!+

n=0

i(1)nz2n+1

(2n + 1)!)

=n=0

(1)nz2n(2n)!

= cos z.

2. Show cos z is an unbounded function in the complex plane. ANSWERS:

Let z = iy, y real. Then limy cos(iy) = limy ey+ey

2= +.

3. Solve z6 = 1; solve z6 = 1.ANSWERS:

Let z = ei. Then ei6 = ei2n, n = 0, 1, . . . , 5. So, z = ein

3 , n = 0, 1, . . . , 5.

Let z = ei. Then ei6 = ei(2n+1), n = 0, 1, . . . , 5. So, z = ei(2n+1)

6 , n =0, 1, . . . , 5.

4. Let D be a domain in the complex plane that is symmetric about thereal axis; that is, if z D, then z D. Assume f(z) is analytic on D and

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if z is real, then f(z) is real. Define g(z) = f(z). Use the Cauchy Riemann

equations and show that g is analytic on D.

ANSWERS: z = x + iy, f(z) = u(x, y) + iv(x, y) where u, v denote real andimaginary parts of f. g(z) = U(x, y) + iV(x, y) where U, V denote real andimaginary parts of g; U(x, y) = u(x, y), V(x, y) = v(x, y). It is an easyapplication of the chain rule to show that U, V satisfy the Cauchy-Riemannequations.

Ux = ux = vy = Vy; Uy = uy = vx = Vx.

5. Find the power series expansion of f(z) about z = z0 for each of thefollowing: QUESTIONS and ANSWERS:

f(z) = ez, z0 = 1; f(z) = eez1 = e

n=0

(z 1)nn!

, |z 1| < .

f(z) = ez2

, z0 = 0; f(z) =n=0

(z2)n

n!=

n=0

(z)2n

n!, |z| < .

f(z) =sin z

z

, z0 = 0; f(z) =1

z

n=0

(

1)n

z2n+1

(2n + 1)!

, =

n=0

(

1)n

z2n

(2n + 1)!

,

|z

| 1,

| 0

iRei

dR4ei4 + 1

| RR4 1 0

as R . Thus,

2

2= lim

R

RR

dx

x4 + 1.

7 page 51Calculate the residue of f(z) = z

2

(z2+a2)(z2+b2)2at each of ai and bi.

At z0 = ai, the residue isz2

(z+ai)(z2+b2)2|z=ai = a2i(b+a)2(ba)2 .

At z0 = bi, apply 1.8.10 with n = 2 and calculate

(d

dz

z2

(z2 + a2)(z + bi)2)|z=bi = a

2 + b2

4bi(b + a)2(b a)2 .

Now just calculate 2i times the sum of the residues.

9 page 52.This one might be a little easier to see with a specific number put in for a.I refer you to a couple examples found athttp://en.wikipedia.org/wiki/Methods of contour integration#Example .28III.29 .E2.80.93 trigonometric integralsOnce you make the z

substitution you are essentially calculating

|z|=1

iz

z4 (4a + 2)z2 + 1 dz =|z|=1

iz

(z2 z0)(z2 z1) dz

where z0 = (2a + 1) 2

a(a + 1) and z1 = (2a + 1) + 2

a(a + 1). Thefunction you are integrating then has simple poles at

z0, z0, z1, z1.

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Only

z0,

z0 are within the interior of the contour. Lets calculate one

of the residues. The residue of f at z = z0 is

z0

(z0 z1)(z0 z0 ) =1

8

a(a + 1).

The residue of f at z = z0 will be the same value.13. Evaluate Res(f(z); z0) for each of the following:

i)f(z) = z4+1

z2(z2)3 ; z0 = 2. Apply 1.8.10 with n = 3. (z 2)3 z4+1

z2(z2)3 = z2 + z2.

So,1

2 limz2d2

dz2 (z2

+ z2

) = 1 +3

16 .

ii) f(z) = z4+1

z2(z2)3 ; z0 = 0. Apply 1.8.10 with n = 2. z2 z4+1z2(z2)3 =

z4+1(z2)3 . So,

limz0

d

dz

z4 + 1

(z 2)3 = 3

16.

iii) f(z) = z4+1

z2(z2)3 ; z0 = 1. f is analytic in a neighborhood of z0 = 1 and so,Res(f(z); z0 = 1) = 0.

14. Study Example 1.10.3 beginning on page 57 and provide the details to

calculate sin(x)x dx.

Let C denote the boundary of the region, {z = rei : r1 < r < R, 0 < < }.By the Cauchy-Goursat Theorem

C

eiz

zdz = 0

since eiz

zhas only a simple pole at z = 0 which not in the region. Parameterize

each of the four subboundaries of C and write

0 =Rr1

eix

x dx +0 ie

iRei

d +r1R

eix

x dx +Cr1

eiz

z dz,

where Cr1 denotes the smaller semi-circular boundary. NoteRr1

sin x

xdx =

1

2i

Rr1

eix eixx

dx.

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17.

f(t) =

0, < t < 0,1, 0 < t < .

Find the corresponding Fourier series.

ANSWER 12

+ 2

n=1

sin(2n1)t2n1

18. Assume f is piecewise continuous on [L, L] and assume f is 2L periodic.Let x0 R. Showx0+2L

x0

f(t)cosnt

Ldt =

LL

f(t)cosnt

Ldt.

f(t)cos ntL

is 2L periodic, so just assume g is 2L periodic and show

x0+2Lx0

g(t)dt =

LL

g(t)dt.

LL

g(t)dt =

x0L

g(t)dt +

Lx0

g(t)dt =

x0L

g(t + 2L)dt +

Lx0

g(t)dt

= x0+2L

L

g(u2L)dt+L

x0

g(t)dt = x0+2L

L

g(u)du+L

x0

g(t)dt = x0+2L

x0

g(t)dt.

19. Given 3(a) p 189, verify 3(b) p 189.

By Parsevals equality,

322

6

n=1

1

(2n 1)6 =1

2

22

f2dt =

20

f2dt

since f2 is even.

20

f2

dt =20

(t4

4t3

+ 4t2

)dt =32

30 .

20. Given f(t) = t, 0 < t < , find the corresponding Fourier cosine seriesand then calculate

n=1

1(2n1)2 . You will evaluate the Fourier cosine series

at some value t0. What theorems or theory do you use so you can performthat evaluation?

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t = 2

4

n=1

cos(2n 1)t(2n 1)2 , 0 < t < .

The RHS is even so,

|t| = 2

4

n=1

cos(2n 1)t(2n 1)2 , < t < 0 0 < t < .

|t| is piecewise continuous on (, ) and so, the Fourier series convergesto the mean value at each t. Since |t| is continuous at 0, the Fourier seriesconverges to

|t

|at t = 0. So, evaluate the RHS at t = 0 and obtain

n=1

1

(2n 1)2 =2

8.

21. Assume f is piecewise continuous on (L, L) and assume f is odd. Show

2

L

L0

f(t)sinnt

Ldt =

1

L

LL

f(t)sinnt

Ldt.

L

Lf(t)sin

nt

Ldt =

0

Lf(t)sin

nt

Ldt +

L

0

f(t)sinnt

Ldt.

In the first expression set s = t.0L

f(t)sinnt

Ldt +

L0

f(t)sinnt

Ldt =

0L

f(s)sin n(s)L

(ds) +L0

f(t)sinnt

Ldt =

L0

f(s)sin nsL

ds +L0

f(t)sin ntL

dt = 2L0

f(t)sin ntL

dt.

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