Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

11.1 Cauchy’s Integral FormulaResult 11.1.1 Cauchy’s Integral Formula.. We will evaluate this integral using the Cauchy integralformula.. Exercise 11.10 Use Cauchy-Goursat, the generaliz

11.1 Cauchy’s Integral Formula

Result 11.1.1 Cauchy’s Integral Formula If f (ζ) is analytic in a compact, closed, nected domain D and z is a point in the interior of D then

con-f (z) = 1

ı2π I

∂D

f (ζ)

ζ − z dζ =

1 ı2π X

Here the set of contours {Ck} make up the positively oriented boundary ∂D of the domain

D More generally, we have

f(n)(z) = n!

ı2π I

∂D

f (ζ) (ζ − z)n+1 dζ = n!

ı2π X

k

I

Ck

f (ζ) (ζ − z)n+1 dζ (11.2)

Cauchy’s Formula shows that the value of f (z) and all its derivatives in a domain are determined by the value of

f (z) on the boundary of the domain Consider the first formula of the result, Equation 11.1 We deform the contour

to a circle of radius δ about the point ζ = z

The remaining integral along Cδ vanishes as δ → 0 because f (ζ) is continuous We demonstrate this with the maximummodulus integral bound The length of the path of integration is 2πδ.

lim

δ→0

I

Cδ

f (ζ) − f (z)

ζ − z dζ

≤ lim

δ→0

(2πδ)1

C

f (ζ)(ζ − z)n+1dζExample 11.1.1 Consider the following integrals where C is the positive contour on the unit circle For the thirdintegral, the point z = −1 is removed from the contour

2 (z−3)(3z−1)1 has singularities at z = 3 and z = 1/3 Since z = 3 is outside the contour, only the singularity at

z = 1/3 will contribute to the value of the integral We will evaluate this integral using the Cauchy integralformula

I

C

1(z − 3)(3z − 1)dz = ı2π



1(1/3 − 3)3



= −ıπ4

3 Since the curve is not closed, we cannot apply the Cauchy integral formula Note that √

z is single-valued andanalytic in the complex plane with a branch cut on the negative real axis Thus we use the Fundamental Theorem

Cauchy’s Inequality Suppose the f (ζ) is analytic in the closed disk |ζ − z| ≤ r By Cauchy’s integral formula,

f(n)(z) = n!

ı2πI

C

f (ζ)(ζ − z)n+1 dζ,

where C is the circle of radius r centered about the point z We use this to obtain an upper bound on the modulus of

C

f (ζ)(ζ − z)n+1 dζ

f c +  eıθ
dθ

If |f (z)| < |f (c)| for any point on |z − c| = , then the continuity of f (z) implies that |f (z)| < |f (c)| in a neighborhood

of that point which would make the value of the integral of |f (z)| strictly less than |f (c)| Thus we conclude that

|f (z)| = |f (c)| for all |z − c| =  Since we can repeat the above procedure for any circle of radius smaller than ,

|f (z)| = |f (c)| for all |z − c| ≤ , i.e all the points in the disk of radius  about z = c are also maxima By recursivelyrepeating this procedure points in this disk, we see that |f (z)| = |f (c)| for all z ∈ D This implies that f (z) is aconstant in the domain By reversing the inequalities in the above method we see that the minimum modulus of f (z)must also occur on the boundary

Result 11.1.6 Extremum Modulus Theorem Let f (z) be analytic in a closed, connected domain, D The extreme values of the modulus of the function must occur on the boundary.

If |f (z)| has an interior extrema, then the function is a constant.

11.2 The Argument Theorem

Result 11.2.1 The Argument Theorem Let f (z) be analytic inside and on C except for isolated poles inside the contour Let f (z) be nonzero on C.

1 ı2π Z

C

f0(z)

f (z) dz = N − P Here N is the number of zeros and P the number of poles, counting multiplicities, of f (z) inside C.

First we will simplify the problem and consider a function f (z) that has one zero or one pole Let f (z) be analyticand nonzero inside and on A except for a zero of order n at z = a Then we can write f (z) = (z − a)ng(z) where g(z)

is analytic and nonzero inside and on A The integral of ff (z)0(z) along A is

1ı2πZ

A

f0(z)

f (z) dz =

1ı2πZ

A

d

dz(log(f (z))) dz

= 1ı2πZ

A

n

z − adz

= n

Now let f (z) be analytic and nonzero inside and on B except for a pole of order p at z = b Then we can write

f (z) = (z−b)g(z)p where g(z) is analytic and nonzero inside and on B The integral of ff (z)0(z) along B is

1ı2πZ

B

f0(z)

f (z) dz =

1ı2πZ

B

d

dz(log(f (z))) dz

= 1ı2πZ

B

−p

z − bdz

= −p

Now consider a function f (z) that is analytic inside an on the contour C except for isolated poles at the points

b1, , bp Let f (z) be nonzero except at the isolated points a1, , an Let the contours Ak, k = 1, , n, be simple,positive contours which contain the zero at ak but no other poles or zeros of f (z) Likewise, let the contours Bk,

k = 1, , p be simple, positive contours which contain the pole at bk but no other poles of zeros of f (z) (SeeFigure 11.1.) By deforming the contour we obtain

A1

B1 B3

B2 A2

Figure 11.1: Deforming the contour C

First note that since |f (z)| > |g(z)| on C, f (z) is nonzero on C The inequality implies that |f (z) + g(z)| > 0

on C so f (z) + g(z) has no zeros on C We well count the number of zeros of f (z) and g(z) using the ArgumentTheorem, (Result11.2.1) The number of zeros N of f (z) inside the contour is

N = 1ı2πI

C

f0(z) + g0(z)

f (z) + g(z) dz

= 1ı2πI

C

f0(z) + f0(z)h(z) + f (z)h0(z)

f (z) + f (z)h(z) dz

= 1ı2πI

C

f0(z)

f (z) dz +

1ı2πI

(Note that since |h(z)| < 1 on C, <(1 + h(z)) > 0 on C and the value of log(1 + h(z)) does not not change intraversing the contour.) This demonstrates that f (z) and f (z) + g(z) have the same number of zeros inside C andproves the result.

ezt

z2(z2+ a2)dz,

where C is any positively oriented contour surrounding the circle |z| = a.

Note that z0 may be either inside or outside of C

Z π 0

ea cos θcos(a sin θ) dθ = π

Exercise 11.10

Use Cauchy-Goursat, the generalized Cauchy integral formula, and suitable extensions to multiply-connected domains

to evaluate the following integrals Be sure to justify your approach in each case

Exercise 11.11

Use Liouville’s theorem to prove the following:

1 If f (z) is entire with <(f (z)) ≤ M for all z then f (z) is constant

2 If f (z) is entire with |f(5)(z)| ≤ M for all z then f (z) is a polynomial of degree at most five

Hint 11.11

Hint 11.12

Hint 11.13

f0(z)

f (z) dz = N − P,where N is the number of zeros and P is the number of poles, (counting multiplicities), of f (z) inside C The theorem

is aptly named, as

1ı2πZ

C

f0(z)

f (z) dz =

1ı2π [log(f (z))]C

= 1ı2π [log |f (z)| + ı arg(f (z))]C

= 12π[arg(f (z))]C.Thus we could write the argument theorem as

1ı2πZ

C

f0(z)

f (z) dz =

12π[arg(f (z))]C = N − P.

Since sin z has a single zero and no poles inside the unit circle, we have

12πarg(sin(z))

C = 1 − 0arg(sin(z))

C = 2πSolution 11.2

1 Since the integrand zsin z2 +5 is analytic inside and on the contour, (the only singularities are at z = ±ı√

5 and atinfinity), the integral is zero by Cauchy’s Theorem

2 First we expand the integrand in partial fractions.

z=−ı

= 12Now we can do the integral with Cauchy’s formula

dz

=Z

C

f (z)(z − α)(z − β)dz

≤ 2πr max

|z|=r

... data-page= "23 ">

-4 -2 4< /p>

-4 -2

24

Figure 11 .2: The contours and the singularities of 3z2< /small>1+1

-6 -4 -2

-6 -4 -2

24 6...

C2

-4 -2 4< /h3>

-4 -2< /h3>

2 4< /h3>

C... deforming C onto C1, C2< /sub> and C3.)

-6 -4 -2 6

-4 -2

Figure