Apply Green's theorem to prove Goursat's theorem
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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 . $$int_T f(z) , dz=int_a^b f(g(t)) g'(t) , dt$$
complex-analysis
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
$endgroup$
add a comment |
$begingroup$
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 . $$int_T f(z) , dz=int_a^b f(g(t)) g'(t) , dt$$
complex-analysis
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
$endgroup$
add a comment |
$begingroup$
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 . $$int_T f(z) , dz=int_a^b f(g(t)) g'(t) , dt$$
complex-analysis
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
$endgroup$
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 . $$int_T f(z) , dz=int_a^b f(g(t)) g'(t) , dt$$
complex-analysis
complex-analysis
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
asked Dec 26 '18 at 10:21
J.GuoJ.Guo
4249
4249
add a comment |
add a comment |
1 Answer
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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 $$int_Cf(z)dz = int_C(Udx-Vdy) + iint_C(Udy + Vdx)$$ And thus by Green's theorem since $int_CW=int_Ddw$ then $$int_DU_ydydx-V_xdxdy + iint_DU_xdxdy + V_ydydx $$$$ int_D(-U_y-V_x)dxdy + iint_D(U_x-V_y)dxdy = 0$$ 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.
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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 $$int_Cf(z)dz = int_C(Udx-Vdy) + iint_C(Udy + Vdx)$$ And thus by Green's theorem since $int_CW=int_Ddw$ then $$int_DU_ydydx-V_xdxdy + iint_DU_xdxdy + V_ydydx $$$$ int_D(-U_y-V_x)dxdy + iint_D(U_x-V_y)dxdy = 0$$ 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.
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$int_Cf(z)dz = int_C(Udx-Vdy) + iint_C(Udy + Vdx)$$ And thus by Green's theorem since $int_CW=int_Ddw$ then $$int_DU_ydydx-V_xdxdy + iint_DU_xdxdy + V_ydydx $$$$ int_D(-U_y-V_x)dxdy + iint_D(U_x-V_y)dxdy = 0$$ 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.
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $$int_Cf(z)dz = int_C(Udx-Vdy) + iint_C(Udy + Vdx)$$ And thus by Green's theorem since $int_CW=int_Ddw$ then $$int_DU_ydydx-V_xdxdy + iint_DU_xdxdy + V_ydydx $$$$ int_D(-U_y-V_x)dxdy + iint_D(U_x-V_y)dxdy = 0$$ 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.
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
$endgroup$
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 $$int_Cf(z)dz = int_C(Udx-Vdy) + iint_C(Udy + Vdx)$$ And thus by Green's theorem since $int_CW=int_Ddw$ then $$int_DU_ydydx-V_xdxdy + iint_DU_xdxdy + V_ydydx $$$$ int_D(-U_y-V_x)dxdy + iint_D(U_x-V_y)dxdy = 0$$ 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.
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
edited Dec 27 '18 at 3:00
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
answered Dec 26 '18 at 23:19
InfinitusInfinitus
885
885
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
Ah I see it . Thanks for explaining !
$endgroup$
– J.Guo
Dec 27 '18 at 17:28
add a comment |
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