MATH529

SP.24: Homework 08

Instructions

The assignment consists solely of problems. The model solutions show verifications through Mathematica. Though beneficial, the Mathematica verifications are not required. While only eight problems need to be formally submitted as homework, it is highly recommended to attempt all questions within sections 17.6-9.

Note: In subsequent lessons on integration in ℂ, integrals of the form
∫((f(z))/(g(z)))dz
will be shown to be expressed in terms of the roots of g(z), hence finding the roots of the equation g(z)=0 is an important skill to form in this homework in preparation for complex integration.

Exercises

There are no exercises in this assignment.

Problems

Solve 5 questions of your choice from the following subsections:  17.6:36-50 (p.856), 17.7:16-32 (p.859),  17.8 (p. 862), 17.R:27-40 (p. 862)

Clear["Global`*"];

Model solution 17.6.35

There are no problems in this assignment.

Solve (e^z)=4i.
Solution: Express in polar form as (e^z)=4 (e^(i π/2)) and take logarithms z=(Subscript[log, e]) 4+i(π/2+2k π)=(Subscript[log, e]) 4+i π(2k+1/2)
Verify:

z[k_] = Log[4] + I π (2k+1/2);
Table[Exp[z[k]],{k,-2,2}]

Model solution 17.7.15

Solve sin z=2.
Solution: From sin(x+i y)=sin  x cosh y+i cos x sinh y=2, set equalities of real and imaginary parts:
(1) sin x cosh y=2
(2) cos x sinh y=0

Equation (2) is satisfied for either:
(2a) cos x=0⇒(Subscript[x, k])=((π/2)+k π), k∈ℤ, or
(2b) sinh y=0⇒y=0

In case (2a), equation (1) states: sin (Subscript[x, k])=(((-1))^k)⇒cosh y=2(((-1))^k). Since cosh y>0, k must be even, k=2n.  The equation cosh y=2 has two solutions (Subscript[y, 1,2])=±ArcCosh[2]=±1.31696. The set of solutions from this case is:
{((Subscript[x, 2n]),(Subscript[y, 1])),((Subscript[x, 2n]),(Subscript[y, 2]))}={((2n+(1/2) )π,ArcCosh[2]),(2n+(1/2) )π,-ArcCosh[2]}, n∈ℤ

In case (2b), equation (1)  states: sin x=2 with no solutions since x∈ℝ.

Verify:

z1[n_]=(2n+(1/2) )π+I ArcCosh[2]; z2[n_]=(2n+(1/2) )π-I ArcCosh[2];
Table[{Sin[z1[n]],Sin[z2[n]]},{n,-2,2}]

Note that the solution can be expressed in various forms:

Solve[Sin[z]==2,z]

ComplexExpand[%]

N[%]

Model solution 17.8.1

Find all values of (sin^(-1))(-i).
Solution: From Apply identity w=(sin^(-1))(z)=-i ln[i z+(Sqrt[1-(z^2)])], to obtain for z=-i
w=-i ln[1+(Sqrt[2])] . The complex square root (Sqrt[2]) has two branches leading to:
(1) (Subscript[w, 1,k])=-i ln[1+(Sqrt[2])] =-i[(Subscript[log, e])(1+(Sqrt[2]))+2k π i]
(2) (Subscript[w, 2,k])=-i ln[1-(Sqrt[2])] =-i ln[r (e^(i θ))]=-i [(Subscript[log, e]) r+i (θ+2k π)] with
1-(Sqrt[2])=r (e^(i θ))⇒r=(Sqrt[2])-1, θ=π, hence
(Subscript[w, 2,k])=-i [(Subscript[log, e])((Sqrt[2])-1)+i π (2k +1)]

Verify:

w1[k_]=-I(Log[1+(Sqrt[2])]+2k π I); w2[k_]=-I(Log[(Sqrt[2])-1]+(2k+1) π I);
Simplify[Table[ComplexExpand[{Sin[w1[k]],Sin[w2[k]]}],{k,-2,2}]]

{N[w1[1]],N[w2[1]]}

In ℝ the roots of sin w=-1 are familiar, (Subscript[w, k])=arcsin(-1)=2kπ+3 π/2. It is useful to build a comparable familiarity with the roots of sin w=-i

p1=ListPlot[Table[{Re[w1[k]],Im[w1[k]]},{k,-4,4}],PlotStyle->Red,Frame->True,FrameLabel->{Re[z],Im[z]},GridLines->{Table[k π,{k,-8,8}],Table[k ,{k,-1,1,.2}]}];
p2=ListPlot[Table[{Re[w2[k]],Im[w2[k]]},{k,-4,4}],PlotStyle->Blue,Frame->True,FrameLabel->{Re[z],Im[z]},GridLines->{Table[k π,{k,-8,8}],Table[k,{k,-1,1,.2}]}];
Show[{p1,p2},PlotRange->{All,{-1,1}},PlotLabel->"Roots of sin(w)=-I"]

Projects

There are no projects in this assignment.