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单词 ClosurePropertiesOfCauchyRiemannEquations
释义

closure properties of Cauchy-Riemann equations


The set of solutions of the Cauchy-Riemann equationsMathworldPlanetmath is closed undera surprisingly large number of operations. For convenience, letus introduce the notational conventions that f and g are complexfunctions with f(x+iy)=u(x,y)+iv(x,y) andg(x+iy)=p(x,y)+iq(x,y). Let D and D denoteopen subsets of the complex planeMathworldPlanetmath.

Theorem 1.

If f:DC andg:DCsatisfy the Cauchy-Riemann equations, so does f+g. Furthermore, if zC, then zf satisfies the Cauchy-Riemann equations.

Proof.

This is an immediate consequence of the linearity of derivatives.∎

Theorem 2.

If f:DC andg:DC satisfy theCauchy-Riemann equations, so does fg.

Proof.

Letting h and k denote the real and imaginary parts of fgrespectively, we have

hx-ky=x(up-vq)-y(uq+vp)
=upx+pux-vqx-qvx-uqy-quy-vpy-pvy
=u(px-qy)-v(py+qx)+p(ux-vy)-q(uy+vx)=0

and

hy+kx=y(up-vq)+x(uq+vp)
=upy+puy-vqy-qvy+uqx+qux+vpx+pvx
=u(py+qx)+v(px-qy)+p(uy+vx)+q(ux-vy)=0.

Theorem 3.

If f:DD andg:DC satisfy theCauchy-Riemann equations, so does fg.

Proof.

Letting h and k denote the real and imaginary parts of fgrespectively, we have

hx-ky=xu(p(x,y),q(x,y))-yv(p(x,y),q(x,y))
=uppx+uqqx-vppy-vqqy
=up(px-qy)+qy(up-vq)+uq(py+qx)-py(uq+vp)=0

and

hy+kx=yu(p(x,y),q(x,y))+xv(p(x,y),q(x,y))
=uppy+uqqy+vppx+vqqx
=up(py+qx)-qx(up-vq)-uq(px-qy)+px(uq+vp)=0

Theorem 4.

If f:DC satisfies the Cauchy-Riemann equations,and has non-vanishing Jacobian, then f-1also satisfies the Cauchy-Riemann equations.

Proof.

Let us denote the real and imaginary parts of f-1 as h and k, respectively.Then, by definition of inverse function, we have

u(h(x,y),k(x,y))=x
v(h(x,y),k(x,y))=y.

Taking derivatives,

uhhx+ukkx=1
uhhy+ukky=0
vhhx+vkkx=0
vhhy+vkky=1

By the Cauchy-Riemann equations, u/h=v/k and u/k=-v/h. Using these relationsMathworldPlanetmath to re-express thederivatives of u as derivatives of v, then subtracting thefourth equation form the first equation and adding the second andthird equations, we obtain

uh(hx-ky)+uk(hy+kx)=0
uh(hy+kx)-uk(hx-ky)=0.

With a little algebraic manipulation, we may conclude

((uh)2+(uk)2)(hy+kx)=0
((uh)2+(uk)2)(hx-ky)=0.

Note that, by the Cauchy-Riemann equations, the Jacobian of fequals the common prefactor of these equations:

(u,v)(h,k)=uhvk-ukvh=(uh)2+(uk)2

Hence, by assumptionsPlanetmathPlanetmath, this quantity differs from zero and we maycancel it to obtain the Cauchy-Riemann equations for f-1.∎

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