Let UC be an open subset, and f:UC be a holomorphic function. With our latest version of Cauchy's Theorem, we have two nice applications.

Application 1:
Let α,β:[t0,t1]U be two non-intersecting C1 paths. Suppose in addition, 1;;that the line segment between α(t) and β(t) for t[t0,t1] is contained in U. Then we can define a linear homotopy ϕ:[t0,t1]×[0,1]U, which is a C1 map. It is namely the map ϕ(t,s)=(1s)α(t)+sβ(t)

Now notice that the boundary of the rectangle can be given by the following curves: α,ϕ(t1,s),β,ϕ(t0,s). We choose let γ run along these curves, and we apply Cauchy's theorem to get $$
\oint_{\alpha}f(z)dz-\oint_{\beta}f(z)dz=\oint_{\phi(t_{0},-)}f(z)dz-\oint_{\phi(t_{1},s)}f(z)dz

Application2:Supposethat$α(t0)=β(t0)$and$α(t1)=β(t1)$.Thenusingtheaboveintegrals,weareleftwith==1;;$αf(z)dz=βf(z)dz$==Application3:(Annulus)If$α(t0)=α(t1)$and$β(t0)=β(t1)$,thenwehaveCauchystheoremforanannulus.Thatis,==1;;$αf(z)dz=bf(z)dz$==.Thisstatementsaysthatunderreasonableconditions,itdoesntmatterwhatringaroundapointyouintegrate.
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