If f(z)Is analytic at alla points inside and on a simple closed curve C then \[\int_{c}^{} f(z)dz=0\]
Proof:Let f(z)=u+iv
\[\int_{c}^{}f(z)dz=\int_{c}^{}(u+iv)(dx+i dy)\]
ie., \[\int_{c}^{}f(z)dz=\int_{c}^{}(udx-vdy)+ i\int_{c}^{}(vdx+udy)\]
We have Green's theorem in a stating that if M(x,y)andN(x,y)are two real value functions having continuous first order partial derivatives in a region R bounded by the curve C,then
\[\int_{c}^{}Mdx+Ndy= \iint_{R}^{}\left (\frac{\partial{ N} }{\partial x} -\frac{\partial{M } }{\partial y} \right ) dx dy\]
Applying this theorem to the two line integrals in the RHS of(1)we obtained
\[\int_{c}^{}f(z)dz= \iint_{R}^{}\left (\frac{-\partial{ v} }{\partial x} -\frac{\partial{u} }{\partial y} \right ) dx dy+ i \iint_{R}^{}\left ( \frac{\partial{u } }{\partial x}-\frac{\partial{ } }{\partial y} \right ) dxdy\]
Since f(z) is analytic, we have Cauchy-Riemann equation:
\[\frac{\partial{u } }{\partial x} =\frac{\partial{ v} }{\partial y} ,\frac{\partial{v } }{\partial x} =\frac{-\partial{ u} }{\partial y} \] and hence we have
\[\int_{c}^{}f(z)dz= \iint_{R}^{}\left ( \frac{\partial{ u} }{\partial y} -\frac{\partial{ u} }{\partial y} \right ) dxdy+i \iint_{R}^{}\left ( \frac{\partial{v } }{\partial y} -\frac{\partial{ v} }{\partial y} \right ) dxdy\]
Thus we get \[\int_{c}^{}f(z)dz=0\]
This proves Cauchy's theorem
Property: If \[ ^{}C_{1}, ^{}C_{2}\] are two simple closed curves such that \[ ^{}C_{2}\]lies entirely within \[ ^{}C_{1}\] and f(z) is analytic \[ ^{}C_{1}, ^{}C_{2}\] and in the region bounded by \[ ^{}C_{1}, ^{}C_{2}\] (known as annual or region)
\[\int_{ ^{}C_{1}}^{}f(z)dz=\int_{ ^{}C_{2}}^{}f(z)dz\]
Proof:
Let us introduce a cross-cut in the form a line segment PQ with the point Pon\[ ^{}C_{1}\] and Q on \[ ^{}C_{2}\].Then the curve PRSPQTUQP as shown in the figure is a simple closed curve and f(z) is analytic inside and on the boundary of C.
Hence by the Cauchy's theorem \[\int_{c}^{}f(z)dz=0\]
Since C is the union of the arts PRSP, PQ, QTUQ,QP, The theorem becomes
\[\int_{ ^{}C_{1}}^{}f(z)dz+\int_{PQ}^{}f(z)dz+\int_{ ^{}-C_{2}}^{}f(z)dz+\int_{QP}^{}f(z)dz=0\]
i.e.,\[\int_{ ^{}C_{1}}^{}f(z)dz+\int_{PQ}^{}f(z)dz-\int_{ ^{}-C_{2}}^{}f(z)dz-\int_{PQ}^{}f(z)dz=0\]
Thus, \[\int_{ ^{}C_{1}}^{} f(z)dz=\int_{ ^{}C_{2}}^{}f(z)dz\]
