29 Jul 2025

Complex Integration and the Cauchy-Goursat Results

Contour parametrization, connected regions, Cauchy-Goursat theorem, Cauchy's integral formula, and Cauchy's inequality.

bsc semester-ii mathematical-physics complex-integration cauchy-integral-formula

Let a directed contour $C$ be parametrized by

\[z=z(t),\qquad a\le t\le b.\]

Because $dz=z’(t)dt$, define

\[\boxed{ \int_C f(z)\,dz =\int_a^b f(z(t))z'(t)\,dt}.\]

Reversing the direction changes the sign. If $F’(z)=f(z)$ throughout a domain containing the contour, the chain rule gives

\[\int_C f(z)dz =\int_a^b\frac{d}{dt}F(z(t))dt =F(z_b)-F(z_a).\]

Simply and multiply connected regions

A region is simply connected if every closed contour in it can be continuously contracted to a point without leaving the region. A region with a hole is multiply connected.

For the unit circle $z=e^{it}$,

\[\oint_{\lvert z\rvert=1}\frac{dz}{z} =\int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt =2\pi i.\]

The integrand is analytic in the punctured plane, but the contour surrounds its missing point $z=0$. It cannot be contracted to a point while remaining in that domain.

A closed contour contracts to a point in a simply connected region. In a multiply connected region, a contour surrounding a hole cannot contract without crossing the excluded set.

Cauchy-Goursat theorem

If $f$ is analytic throughout a simply connected region and on its boundary $C$, then

\[\boxed{\oint_C f(z)\,dz=0}.\]

Under continuous first partial derivatives, the cancellation follows directly from Green’s theorem. With $f=u+iv$ and $dz=dx+i\,dy$,

\[f\,dz=(u\,dx-v\,dy)+i(v\,dx+u\,dy).\]

For counterclockwise $C=\partial D$, Green’s theorem gives

\[\oint_C(u\,dx-v\,dy) =\iint_D(-v_x-u_y)dA,\]

and

\[\oint_C(v\,dx+u\,dy) =\iint_D(u_x-v_y)dA.\]

Both integrands vanish by the Cauchy-Riemann equations. Goursat’s form of the theorem removes the extra assumption that $f’$ is continuous.

Cauchy’s integral formula

Let $f$ be analytic on and inside a positively oriented simple contour $C$, and let $z_0$ lie inside it. Remove a small circle $C_\varepsilon$ about $z_0$. The integrand

\[\frac{f(z)}{z-z_0}\]

is analytic in the region between the two contours, so Cauchy-Goursat gives

\[\oint_C\frac{f(z)}{z-z_0}dz =\oint_{C_\varepsilon}\frac{f(z)}{z-z_0}dz.\]

Split $f(z)=f(z_0)+[f(z)-f(z_0)]$. On

\[z-z_0=\varepsilon e^{it}, \qquad dz=i\varepsilon e^{it}dt,\]

the constant part is

\[f(z_0)\int_0^{2\pi}i\,dt=2\pi i f(z_0).\]

For the remaining part, continuity makes

\[\max_{C_\varepsilon}\lvert f(z)-f(z_0)\rvert\to0.\]

Its integral has magnitude at most this maximum times

\[\frac{\operatorname{length}(C_\varepsilon)}{\varepsilon} =\frac{2\pi\varepsilon}{\varepsilon}=2\pi,\]

so it tends to zero. Therefore

\[\boxed{ f(z_0)=\frac{1}{2\pi i} \oint_C\frac{f(z)}{z-z_0}dz}.\]

Repeated differentiation with respect to $z_0$ gives

\[\boxed{ f^{(n)}(z_0)=\frac{n!}{2\pi i} \oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz}.\]

Cauchy’s inequality

If $\lvert f(z)\rvert\le M$ on the circle $\lvert z-z_0\rvert=R$, then

\[\begin{aligned} \lvert f^{(n)}(z_0)\rvert &\le\frac{n!}{2\pi} \oint_C\frac{\lvert f(z)\rvert}{R^{n+1}}\lvert dz\rvert\\ &\le\frac{n!}{2\pi}\frac{M}{R^{n+1}}(2\pi R). \end{aligned}\]

Hence

\[\boxed{\lvert f^{(n)}(z_0)\rvert\le\frac{n!M}{R^n}}.\]

The unit-circle integral and sample Cauchy-formula integrals are checked in the Unit III Maxima worksheet.