# 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 $D^{\prime}$ denote open subsets of the complex plane.

###### Theorem 1.

If $f\colon D\to\mathbb{C}$ and $g\colon D\to\mathbb{C}$ satisfy the Cauchy-Riemann equations, so does $f+g$. Furthermore, if $z\in\mathbb{C}$, then $zf$ satisfies the Cauchy-Riemann equations.

###### Proof.

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

###### Theorem 2.

If $f\colon D\to\mathbb{C}$ and $g\colon D\to\mathbb{C}$ satisfy the Cauchy-Riemann equations, so does $f\cdot g$.

###### Proof.

Letting $h$ and $k$ denote the real and imaginary parts of $f\cdot g$ respectively, we have

 $\displaystyle{\partial h\over\partial x}-{\partial k\over\partial y}$ $\displaystyle={\partial\over\partial x}\left(up-vq\right)-{\partial\over% \partial y}\left(uq+vp\right)$ $\displaystyle=u{\partial p\over\partial x}+p{\partial u\over\partial x}-v{% \partial q\over\partial x}-q{\partial v\over\partial x}-u{\partial q\over% \partial y}-q{\partial u\over\partial y}-v{\partial p\over\partial y}-p{% \partial v\over\partial y}$ $\displaystyle=u\left({\partial p\over\partial x}-{\partial q\over\partial y}% \right)-v\left({\partial p\over\partial y}+{\partial q\over\partial x}\right)+% p\left({\partial u\over\partial x}-{\partial v\over\partial y}\right)-q\left({% \partial u\over\partial y}+{\partial v\over\partial x}\right)=0$

and

 $\displaystyle{\partial h\over\partial y}+{\partial k\over\partial x}$ $\displaystyle={\partial\over\partial y}\left(up-vq\right)+{\partial\over% \partial x}\left(uq+vp\right)$ $\displaystyle=u{\partial p\over\partial y}+p{\partial u\over\partial y}-v{% \partial q\over\partial y}-q{\partial v\over\partial y}+u{\partial q\over% \partial x}+q{\partial u\over\partial x}+v{\partial p\over\partial x}+p{% \partial v\over\partial x}$ $\displaystyle=u\left({\partial p\over\partial y}+{\partial q\over\partial x}% \right)+v\left({\partial p\over\partial x}-{\partial q\over\partial y}\right)+% p\left({\partial u\over\partial y}+{\partial v\over\partial x}\right)+q\left({% \partial u\over\partial x}-{\partial v\over\partial y}\right)=0.$

###### Theorem 3.

If $f\colon D\to D^{\prime}$ and $g\colon D^{\prime}\to\mathbb{C}$ satisfy the Cauchy-Riemann equations, so does $f\circ g$.

###### Proof.

Letting $h$ and $k$ denote the real and imaginary parts of $f\circ g$ respectively, we have

 $\displaystyle{\partial h\over\partial x}-{\partial k\over\partial y}$ $\displaystyle={\partial\over\partial x}u(p(x,y),q(x,y))-{\partial\over\partial y% }v(p(x,y),q(x,y))$ $\displaystyle={\partial u\over\partial p}{\partial p\over\partial x}+{\partial u% \over\partial q}{\partial q\over\partial x}-{\partial v\over\partial p}{% \partial p\over\partial y}-{\partial v\over\partial q}{\partial q\over\partial y}$ $\displaystyle={\partial u\over\partial p}\left({\partial p\over\partial x}-{% \partial q\over\partial y}\right)+{\partial q\over\partial y}\left({\partial u% \over\partial p}-{\partial v\over\partial q}\right)+{\partial u\over\partial q% }\left({\partial p\over\partial y}+{\partial q\over\partial x}\right)-{% \partial p\over\partial y}\left({\partial u\over\partial q}+{\partial v\over% \partial p}\right)=0$

and

 $\displaystyle{\partial h\over\partial y}+{\partial k\over\partial x}$ $\displaystyle={\partial\over\partial y}u(p(x,y),q(x,y))+{\partial\over\partial x% }v(p(x,y),q(x,y))$ $\displaystyle={\partial u\over\partial p}{\partial p\over\partial y}+{\partial u% \over\partial q}{\partial q\over\partial y}+{\partial v\over\partial p}{% \partial p\over\partial x}+{\partial v\over\partial q}{\partial q\over\partial x}$ $\displaystyle={\partial u\over\partial p}\left({\partial p\over\partial y}+{% \partial q\over\partial x}\right)-{\partial q\over\partial x}\left({\partial u% \over\partial p}-{\partial v\over\partial q}\right)-{\partial u\over\partial q% }\left({\partial p\over\partial x}-{\partial q\over\partial y}\right)+{% \partial p\over\partial x}\left({\partial u\over\partial q}+{\partial v\over% \partial p}\right)=0$

###### Theorem 4.

If $f\colon D\to\mathbb{C}$ 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

 $\displaystyle u(h(x,y),k(x,y))$ $\displaystyle=x$ $\displaystyle v(h(x,y),k(x,y))$ $\displaystyle=y.$

Taking derivatives,

 $\displaystyle{\partial u\over\partial h}{\partial h\over\partial x}+{\partial u% \over\partial k}{\partial k\over\partial x}$ $\displaystyle=1$ $\displaystyle{\partial u\over\partial h}{\partial h\over\partial y}+{\partial u% \over\partial k}{\partial k\over\partial y}$ $\displaystyle=0$ $\displaystyle{\partial v\over\partial h}{\partial h\over\partial x}+{\partial v% \over\partial k}{\partial k\over\partial x}$ $\displaystyle=0$ $\displaystyle{\partial v\over\partial h}{\partial h\over\partial y}+{\partial v% \over\partial k}{\partial k\over\partial y}$ $\displaystyle=1$

By the Cauchy-Riemann equations, $\partial u/\partial h=\partial v/\partial k$ and $\partial u/\partial k=-\partial v/\partial h$. 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

 $\displaystyle{\partial u\over\partial h}\left({\partial h\over\partial x}-{% \partial k\over\partial y}\right)+{\partial u\over\partial k}\left({\partial h% \over\partial y}+{\partial k\over\partial x}\right)$ $\displaystyle=0$ $\displaystyle{\partial u\over\partial h}\left({\partial h\over\partial y}+{% \partial k\over\partial x}\right)-{\partial u\over\partial k}\left({\partial h% \over\partial x}-{\partial k\over\partial y}\right)$ $\displaystyle=0.$

With a little algebraic manipulation, we may conclude

 $\displaystyle\left(\left({\partial u\over\partial h}\right)^{2}+\left({% \partial u\over\partial k}\right)^{2}\right)\left({\partial h\over\partial y}+% {\partial k\over\partial x}\right)$ $\displaystyle=0$ $\displaystyle\left(\left({\partial u\over\partial h}\right)^{2}+\left({% \partial u\over\partial k}\right)^{2}\right)\left({\partial h\over\partial x}-% {\partial k\over\partial y}\right)$ $\displaystyle=0.$

Note that, by the Cauchy-Riemann equations, the Jacobian of $f$ equals the common prefactor of these equations:

 ${\partial(u,v)\over\partial(h,k)}={\partial u\over\partial h}{\partial v\over% \partial k}-{\partial u\over\partial k}{\partial v\over\partial h}=\left({% \partial u\over\partial h}\right)^{2}+\left({\partial u\over\partial k}\right)% ^{2}$

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 ClosurePropertiesOfCauchyRiemannEquations 2013-03-22 17:44:20 2013-03-22 17:44:20 rspuzio (6075) rspuzio (6075) 14 rspuzio (6075) Theorem msc 30E99 TangentialCauchyRiemannComplexOfCinftySmoothForms ACRcomplex