closure properties of Cauchy-Riemann equations
The set of solutions of the Cauchy-Riemann equations is closed under
a surprisingly large number of operations. For convenience, let
us introduce the notational conventions that f and g are complex
functions with f(x+iy)=u(x,y)+iv(x,y) and
g(x+iy)=p(x,y)+iq(x,y). Let D and D′ denote
open subsets of the complex plane
.
Theorem 1.
If f:D→C and g:D→C satisfy the Cauchy-Riemann equations, so does f+g. Furthermore, if z∈C, then zf satisfies the Cauchy-Riemann equations.
Proof.
This is an immediate consequence of the linearity of derivatives. ∎
Theorem 2.
If f:D→C and g:D→C satisfy the Cauchy-Riemann equations, so does f⋅g.
Proof.
Letting h and k denote the real and imaginary parts of f⋅g respectively, we have
∂h∂x-∂k∂y | =∂∂x(up-vq)-∂∂y(uq+vp) | ||
=u∂p∂x+p∂u∂x-v∂q∂x-q∂v∂x-u∂q∂y-q∂u∂y-v∂p∂y-p∂v∂y | |||
=u(∂p∂x-∂q∂y)-v(∂p∂y+∂q∂x)+p(∂u∂x-∂v∂y)-q(∂u∂y+∂v∂x)=0 |
and
∂h∂y+∂k∂x | =∂∂y(up-vq)+∂∂x(uq+vp) | ||
=u∂p∂y+p∂u∂y-v∂q∂y-q∂v∂y+u∂q∂x+q∂u∂x+v∂p∂x+p∂v∂x | |||
=u(∂p∂y+∂q∂x)+v(∂p∂x-∂q∂y)+p(∂u∂y+∂v∂x)+q(∂u∂x-∂v∂y)=0. |
∎
Theorem 3.
If f:D→D′ and g:D′→C satisfy the Cauchy-Riemann equations, so does f∘g.
Proof.
Letting h and k denote the real and imaginary parts of f∘g respectively, we have
∂h∂x-∂k∂y | =∂∂xu(p(x,y),q(x,y))-∂∂yv(p(x,y),q(x,y)) | ||
=∂u∂p∂p∂x+∂u∂q∂q∂x-∂v∂p∂p∂y-∂v∂q∂q∂y | |||
=∂u∂p(∂p∂x-∂q∂y)+∂q∂y(∂u∂p-∂v∂q)+∂u∂q(∂p∂y+∂q∂x)-∂p∂y(∂u∂q+∂v∂p)=0 |
and
∂h∂y+∂k∂x | =∂∂yu(p(x,y),q(x,y))+∂∂xv(p(x,y),q(x,y)) | ||
=∂u∂p∂p∂y+∂u∂q∂q∂y+∂v∂p∂p∂x+∂v∂q∂q∂x | |||
=∂u∂p(∂p∂y+∂q∂x)-∂q∂x(∂u∂p-∂v∂q)-∂u∂q(∂p∂x-∂q∂y)+∂p∂x(∂u∂q+∂v∂p)=0 |
∎
Theorem 4.
If f:D→C satisfies the Cauchy-Riemann equations, and has non-vanishing Jacobian, then f-1 also 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,
∂u∂h∂h∂x+∂u∂k∂k∂x | =1 | ||
∂u∂h∂h∂y+∂u∂k∂k∂y | =0 | ||
∂v∂h∂h∂x+∂v∂k∂k∂x | =0 | ||
∂v∂h∂h∂y+∂v∂k∂k∂y | =1 |
By the Cauchy-Riemann equations, ∂u/∂h=∂v/∂k and ∂u/∂k=-∂v/∂h. Using these relations to re-express the
derivatives of u as derivatives of v, then subtracting the
fourth equation form the first equation and adding the second and
third equations, we obtain
∂u∂h(∂h∂x-∂k∂y)+∂u∂k(∂h∂y+∂k∂x) | =0 | ||
∂u∂h(∂h∂y+∂k∂x)-∂u∂k(∂h∂x-∂k∂y) | =0. |
With a little algebraic manipulation, we may conclude
((∂u∂h)2+(∂u∂k)2)(∂h∂y+∂k∂x) | =0 | ||
((∂u∂h)2+(∂u∂k)2)(∂h∂x-∂k∂y) | =0. |
Note that, by the Cauchy-Riemann equations, the Jacobian of f equals the common prefactor of these equations:
∂(u,v)∂(h,k)=∂u∂h∂v∂k-∂u∂k∂v∂h=(∂u∂h)2+(∂u∂k)2 |
Hence, by assumptions, this quantity differs from zero and we may
cancel it to obtain the Cauchy-Riemann equations for f-1.
∎
Title | closure properties of Cauchy-Riemann equations |
---|---|
Canonical name | ClosurePropertiesOfCauchyRiemannEquations |
Date of creation | 2013-03-22 17:44:20 |
Last modified on | 2013-03-22 17:44:20 |
Owner | rspuzio (6075) |
Last modified by | rspuzio (6075) |
Numerical id | 14 |
Author | rspuzio (6075) |
Entry type | Theorem![]() |
Classification | msc 30E99 |
Related topic | TangentialCauchyRiemannComplexOfCinftySmoothForms |
Related topic | ACRcomplex |