## Cauchy’s Theorem

Cauchy’s Theorem:

Let $R$ be the closed region consisting of all points interior to and on the simple closed contour $C$.

If $f$ is analytic in $R$ and $f'$ is continuous in $R$,

$\int_C f(z)\,dz=0$

(This is the precursor of Cauchy-Goursat Theorem, which allows us to drop the condition that $f'$ is continuous.)

Proof Using Green’s Theorem:

Let $C$ denote a positively oriented simple closed contour $z=z(t)$, $a\leq t\leq b$.

$\int_C f(z)\,d(z)=\int_a^b f[z(t)]z'(t)\,dt$ where $f(z)=u(x,y)+iv(x,y)$ and $z(t)=x(t)+iy(t)$. Thus

\begin{aligned} \int_C f(z)\,dz&=\int_a^b (u+iv)(x'+iy')\,dt\\ &=\int_a^b (ux'-vy')\,dt+i\int_a^b (vx'+uy')\,dt\\ &=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy \end{aligned}

Next, we need Green’s Theorem:

$\int_C P\,dx+Q\,dy=\iint_R (Q_x-P_y)\,dA$

By assumption $f'$ is continuous in $R$, thus the first-order partial derivatives of $u$ and $v$ are also continous. This is exactly what we need for Green’s Theorem.

Continuing from above, we get $\int_C f(z)\,dz=\iint_R (-v_x-u_y)\,dA+i\iint_R(u_x-v_y)\,dA$ which is exactly zero in view of the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$!