MATH529

SP.22: Homework 14

Instructions

The assignment consists solely of exercises. The model solutions carry out tedious differentiations and other elementary calculations in Mathematica. The built-in Mathematica residue function is also used to verify solutions.  Feel free to use these facilities if you are confident in your differentiation abilities.

Exercises

19.5, p.913: 1-30, 19.6 p.919:1-30

Model solution 19.5.31

Compute
(∫_(|z-3|=1)^□)f(z)dz=(∫_(|z-3|=1)^□)(((e^(i z))+sin z)/(((z-π))^4))dz

Solution:  The integrand f(z) has a pole of order 4 at z=π within the circle |z-3|=1
Apply residue theorem

(∫_(|z-3|=1)^□)f(z)dz=(1/(3!))2π i (lim_(z→π))((d^3)/(d(z^3)))[(((z-π))^4)(((e^(i z))+sin z)/(((z-π))^4))]=(1/3)π i (lim_(z→π))( -((ie)^(i z)) -cos z)=(1/3)π i(i+1)=(1/3)π(i-1)

f[z_]=(((E^(I z))+Sin[z])/(((z-π))^4)); ((2π I)/(3!)) D[(((z-π))^4) f[z],{z,3}] /. z->π

2 π I Residue[f[z],{z,π}]

Model solution 19.5.32

Compute
(∫_(|z-1|=1)^□)f(z)dz=(∫_(|z-1|=1)^□)((cos z)/((((z-1))^2)((z^2)+9)))dz

Solution:  The integrand f(z) has a pole of order 2 at z=1 within the circle |z-1|=1
Apply residue theorem

(∫_(|z-3|=1)^□)f(z)dz=(1/(1!))2π i (lim_(z→1))(d/(dz))[(((z-1))^2)((cos z)/((((z-1))^2)((z^2)+9)))]=2π i (lim_(z→1))((-((z^2)+9)sin z -2z cos z )/((((z^2)+9))^2))=2π i ((-10 sin 1-2 cos 1)/100)=-π i ((5 sin 1+ cos 1)/25)

f[z_]=((Cos[z])/((((z-1))^2)((z^2)+9))); 2 π I Residue[f[z],{z,1}]

Model solution 19.6.21

Compute
I=(∫_(-∞)^∞)f(x)dx=(∫_(-∞)^∞)((cos x)/((x^2)+1))dx

Solution:  Consider the contour C shown below formed by a line segment and semicircle in the upper complex plane.  (Note: Mathematica graphics instructions are shown for pedagogical purposes. Feel free to include such figures in your answers, but they are not required).

R=2;
fig1 = Graphics[{Line[{{-R,0},{R,0}}],Arrow[{{-R/2,0},{R/2,0}}],Circle[{0,0},R,{0,π}],
Point[{0,1}]
}];
Show[fig1,Frame->True,FrameLabel->{"x","y"},GridLines->Automatic,PlotLabel->"Integration contour in ℂ"]

Consider the complex integral on C of f(z)=cos z/((z^2)+1) and separate into the integral on the line segment and the semicircle S,  z(θ)=R (e^(i θ)), dz=i R (e^(i θ)) dθ ,

J=(∫_C)f(z)dz=(Subscript[I, 1])+(Subscript[I, 2])=(∫_(-R)^R)f(x)dx+(∫_S)f(z)dz

The Bounding Theorem 19.6.2 (cf. Lesson 24) states that for g(z)=P(z)/Q(z), a rational function with degree n of polynomial P(z) and  degree of  m⩾n+1, and a>0,

(lim_(R⟶∞))(∫_S) g(z) (e^(i a z))dz=0, (lim_(R⟶∞))(∫_S) g(z) cos(a z)dz=0, (lim_(R⟶∞))(∫_S) g(z) sin(a z)dz=0

Apply to f(z)=cos z/((z^2)+1), (i.e., P(z)=1, Q(z)=(z^2)+1 of degree m=2⩾0+1), to obtain (lim_(R⟶∞))(Subscript[I, 2])=0, hence

I=(lim_(R→∞))(∫_C)f(z)dz=2π i Res{f(z),i}

Evaluate the residue at the simple pole (Subscript[z, 0])=i,

Res{f(z),i}=(lim_(z→i)) (z-i) ((cos z)/((z^2)+1))=((cos i)/(2i)) , to obtain

I=2π i ((cos i)/(2i))=π  cos i=π cosh 1.

Verify in Mathematica

f[z_]=((Cos[z])/((z^2)+1)); 2 π I Residue[f[z],{z,I}]

Problems

There are no problems in this assignment.

Projects

There are no projects in this assignment.