# Apply Green's theorem to prove Goursat's theorem

Suppose $$f$$ is continuously complex differentiable on $$\Omega$$ , and $$T \subset \Omega$$ is a triangle whose interior is also contained in $$\Omega$$ . Apply Green's theorem to show that $$\int_T f(z) \, dz=0$$

This is an exercise in Stein's complex analysis Page$$65$$ .
My attempt:
Let $$f(z)=u(x,y)+iv(x,y)$$ , $$dz=dx+idy$$ then apply Green's theorem I can get the desire conclusion . But I can not prove that $$dz=dx+idy$$ , since the definition of integral along a curve only has one variable . $$\int_T f(z) \, dz=\int_a^b f(g(t)) g'(t) \, dt$$

Let us denote a function $$f(z)$$ by $$f(z)= u(x,y) + iv(x,y)$$where $$u,v$$ are continuous upon $$w,x,t,y$$ which can then be shown that $$f(z)dz = (u+iv)(dx+idy)=(udx-vdy)+i(udy+vdx)$$ Using Green's theorem $$\int_CP(x,y)dx+Q(x,y)dy=\int_DQ_xP_ydxdy$$ we can substitute $$f(z)$$ in which then yields $$\int_Cf(z)dz = \int_C(Udx-Vdy) + i\int_C(Udy + Vdx)$$ And thus by Green's theorem since $$\int_CW=\int_Ddw$$ then $$\int_DU_ydydx-V_xdxdy + i\int_DU_xdxdy + V_ydydx$$$$\int_D(-U_y-V_x)dxdy + i\int_D(U_x-V_y)dxdy = 0$$ since $$f'(z)$$ exists, the function is analytic, so by the Cauchy Reimann Equations, $$U_x = V_y$$ and $$U_y = -V_x$$, implying that all integrals evaluate to 0.
• Very appreciate for your help, but I think my question is why we can write $dz=dx + idy$ , which seems obvious . – J.Guo Dec 27 '18 at 4:03
• Since the imaginary axis is the y axis and the real axis is the x axis $dz = dx + idy = (x’(t) + y’(t)i)dt = z’(t)dt$ holds. So $$\int_C f(z)dz = \int_a^b f(z(t))z’(t)dt$$ – Infinitus Dec 27 '18 at 17:17