Week 4 Worksheet - Solutions

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Contours

  1. We calculate γ(t)=3+i so that |γ(t)|=10. Then (γ)=3510dt=210

  2. The following parameterization works. γ1(t)=2t on [0,1]γ2(t)=2+3ti on [0,1]γ3(t)=3i+2(1t) on [0,1]γ4(t)=3i(1t) on [0,1]

    1. A straight line from i to 2.

    2. The reverse of Γ=(γ1,γ2,γ3) should be (γ3~,γ2~,γ1~).

Contour Integration

    1. We take γ(t)=t+it on [0,1] so that x=y=t and γ(t)=1+i. Thus the integral is 01(tt+it2)(1+i)dt=(i1)01t2dt=i13
    2. Here γ(t)=it on [0,1] so x=0 and y=t. Also γ(t)=i. 01(t)idt=i2
    3. Use γ(t)=i+t on [0,1]. Here x=t and y=1 and γ(t)=1. 01(t1+it2)dt=121+i3
    1. γ1(t)=2ieit and γ1(t)2=2eit 02π12eit2ieitdt=02πidt=2πi
    2. γ2(t)=ieit and (γ2(t)i)3=e3it 0π/21e3it(ieit)dt=i0π/2e2itdt=i(eiπ2i12i)=1
  1. Take γ(t)=1+eit on [0,2π]. Then γ(t)=ieit. |γ(t)|2=(1+eit)(1+eit)=2+eit+eit and the integral is 02π(2+eit+eit)ieitdt=i02π2eit+e2it+1dt=2πi

Antiderivatives

    1. We can get an antiderivative F from integration by parts F(z)=z2sin(z)=z2cos(z)+2zcos(z)=z2cos(z)+2zsin(z)2sin(z)=(2z2)cos(z)+2zsin(z) and the fundamental theorem of contour integration lets us evaluate the contour integral easily. 0if=F(i)F(0)=3cos(i)+2isin(i)2
    2. Integration by parts again helps with the antiderivative F(z)=zexp(iz)=1izexp(iz)1iexp(iz)=izexp(iz)+exp(iz) and 0if=F(i)F(0)=exp(1)+exp(1)1=2e1
    1. Γ1=(γ1,γ2) where γ1(t)=it on [0,1] and γ2(t)=i+t on [0,1]. Then Γ1f=γ1f+γ2f=01t2idt+01(1+t2)dt=i3+1+13=4+i3
    2. Γ2=(γ3,γ4) where γ3(t)=t on [0,1] and γ4(t)=1+it on [0,1]. Then Γ2f=γ3f+γ4f=01t2dt+01(1+t2)idt=13+i+i3=4i+13
    3. There is no possibility that f has an antiderivative. If it did, then the contour integral of f over the closed contour Γ=(γ1,γ2,γ4~,γ3~) would be zero and our answers to a) and b) would have to be equal.
    1. γ(t)=3eit on [0,π]
    2. f(z)=1/z2 has an antiderivative F(z)=1/z on C{0}. Therefore γf=F(γ(π))F(γ(0))=1313=23
    3. In this case γ~(t)=3ei(πt). The function f still has an antiderivative, so γ~f=F(γ~(π))F(γ~(0))=F(3)F(3)=1313=23
    4. The answer to c) is the negative of the answer to b)
  1. The function h(z)=f(z)g(z)+f(z)g(z) has an antiderivative H(z)=f(z)g(z) by the product rule. Therefore γfg+γfg=γfg+fg=γh=H(z1)H(z0)=f(z1)g(z1)f(z0)g(z0) and subtracting the contour integral of fg gives the result.

The Principal Logarithm

    1. Log(z)=ln(|z|)+iArg(z) can take any real value and any imaginary value between π and π. Therefore the range is {x+iy:xR,π<yπ}.
    2. In order to calculate Log(exp(z)) we need the absolute value and the principal argument of exp(z). Write z=x+iy so that exp(z)=ex(cos(y)+isin(y)).
    1. Take a=b=e3πi/4. Then ab=e3πi/2 and Log(e3πi/2)=ln(1)iπ2 because we must use the principal argument, but this is not twice Log(e3πi/4)=ln(1)+i3π4.
    2. Let θ and ψ be the principal arguments of a and b respectively. We know from polar multiplication that θ+ψ is an argument of ab. It may not be the principal argument, but there is nZ such that Arg(ab)=θ+ψ+2πn. Thus Log(a)+Log(b)+2πin=ln(|a|)+iArg(a)+ln(|b|)+iArg(b)+2πin=ln(|ab|)+i(θ+ψ+2πn)=ln(|ab|)+iArg(ab)=Log(ab)
  1. |1+i|=2 and Arg(1+i)=π/4, so Log(1+i)=ln(2)+iπ4.

    |1i|=2 and Arg(1i)=3π/4 so Log(1i)=ln(2)i3π4.

    1. ii=exp(iLog(i))=exp(i(ln(1)+iπ/2))=exp(π/2)
    2. ab+c=exp((b+c)Log(a))=exp(bLog(a))+exp(cLog(a))=abac
    3. (a1/2)2=(exp(12Log(a))2=exp(12Log(a)+12Log(a))=exp(Log(a))=a
    4. Take a=b=ei3π/4. Then a1/2=ei3π/8 but ab=e3πi/2 and (ab)1/2=eiπ/4.