- Suppose \(f\) is continuously complex diferentiable on \(\Omega ,\) and \(T \subset \Omega\) is a triangle whose interior is also contained in Ω. Apply Green’s theorem to show that
\[ \int_ {T} f (z) d z = 0. \]
This provides a proof of Goursat’s theorem under the additional assumption that \(f ^ { \prime }\) is continuous.
[Hint: Green’s theorem says that if \(( F , G )\) is a continuously diferentiable vector field, then
\[ \int_ {T} F d x + G d y = \int_ {\mathrm{Interiorof} T} \left(\frac {\partial G}{\partial x} - \frac {\partial F}{\partial y}\right) d x d y. \]
For appropriate \(F\) and G, one can then use the Cauchy-Riemann equations.]
Solution
Write \[ f=u+iv, \qquad dz=dx+i\,dy. \] Then \[ f(z)\,dz=(u\,dx-v\,dy)+i(v\,dx+u\,dy). \] Because \(f\) is continuously complex differentiable, \(u\) and \(v\) are \(C^1\) and satisfy the Cauchy–Riemann equations \[ u_x=v_y, \qquad u_y=-v_x. \]
Apply Green’s theorem to the real part: \[ \int_T u\,dx-v\,dy =\iint_{\operatorname{int}T}(-v_x-u_y)\,dx\,dy=0. \] Apply it again to the imaginary part: \[ \int_T v\,dx+u\,dy =\iint_{\operatorname{int}T}(u_x-v_y)\,dx\,dy=0. \] Hence both real and imaginary parts vanish, and therefore \[ \boxed{\displaystyle \int_T f(z)\,dz=0}. \]