Let
γ=γ1+γ2+γ3,γ1(t)=eit,t∈[0,2π]γ2(t)=−1+2e−2it,t∈[0,2π]γ3(t)=1−i+eit,t∈[π2,9π2]
Calculate the value of n(γ,z) as z takes its value in C∖γ.
γ is the image of the closed curve.
as γ is a closed and smooth by parts, we have that
n(γ,z)=12πi∫γdcc−z=12πi[∫γ1dcc−z+∫γ2dcc−z+∫γ3dcc−z]
as γ′1(t)=ieit, γ′2(t)=−4ie−2it and γ′3(t)=ieit
i can compute those integrals as
n(γ,z)=12πi[2π∫0ieitdteit−z+2π∫0−4ie−2itdt−1+2e−2it−z+9π/2∫π/2ieit1−i+eit−z]
but when i compute then for any point i am finding 0, did i made any mistake or am i making one when computing those integrals?
Answer
The value depends on z. Note that γ1 runs once counterclockwise around the circle with center 0 and radius 1, γ2 runs twice clockwise around the circle with center −1 and radius 2, γ3 runs twice counterclockwise around the circle with center 1−i and radius 1. Let Di be the open disk bounded by γi. Note that D1⊂D2. Drawing a picture is helpful. For a point z∈C∖γ we get n(γ,z)=n(γ1,z)+n(γ2,z)+n(γ3,z). We have n(γ1,z)=1 for z∈D1, n(γ1,z)=0 for z∉D1, n(γ2,z)=−2 for z∈D2, n(γ2,z)=0 for z∉D2, n(γ3,z)=2 for z∈D3, n(γ3,z)=0 for z∉D3. Thus:
If z∈D1∩D3, then n(γ,z)=1−2+2=1.
If z∈D1∖D3, then n(γ,z)=1−2+0=−1.
If z∈D2∖(D1∪D3), then n(γ,z)=0−2+0=−2.
If z∈(D2∩D3)∖D1, then n(γ,z)=0−2+2=0.
If z∈D3∖D2, then n(γ,z)=0+0+2=2.
If z∉D1∪D2∪D3, then n(γ,z)=0+0+0=0.
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