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.

  • $\begingroup$ Very appreciate for your help, but I think my question is why we can write $dz=dx + idy$ , which seems obvious . $\endgroup$ – J.Guo Dec 27 '18 at 4:03
  • $\begingroup$ 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 $$ \int_C f(z)dz = \int_a^b f(z(t))z’(t)dt$$ $\endgroup$ – Infinitus Dec 27 '18 at 17:17
  • $\begingroup$ Ah I see it . Thanks for explaining ! $\endgroup$ – J.Guo Dec 27 '18 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.