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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 . $$\int_T f(z) \, dz=\int_a^b f(g(t)) g'(t) \, dt$$

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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.

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  • $\begingroup$ 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

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