Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $U(r,\theta),V(r, \theta)$ are continuously differentiable functions on some polar rectangle $R = \{(r, \theta) \colon r \in (a,b), \theta \in (\theta_1, \theta_2) \} \subseteq \mathbb{R}^2.$ Furthermore, assume that $U$ and $V$ satisfy the polar Cauchy-Riemann equations in $R$:

$$rU_r = V_\theta, U_\theta = -rV_r.$$

If we now view $R$ as a subset of $\mathbb{C}$ rather than $\mathbb{R}^2$, we can define the function $f : R \to \mathbb{C}$ by $f(re^{i\theta}) = U(r, \theta) + iV(r,\theta).$ Prove that $f$ is analytic on $R$.

I am linking this problem to a previous post: Proof of Cauchy Riemann Equations in Polar Coordinates. I believe I am asking a similar question. However, to my best knowledge, the answers to the linked post actually establish the converse of my statement above. That is, they show that analyticity of $f$ implies that these polar Cauchy-Riemann equations are satisfied.

Here's what I have so far: I do know that a function $f(x + iy) = U(x,y) + iV(x,y)$ is analytic when its real and imaginary parts are continuously differentiable and satisfy the rectangular Cauchy-Riemann equations $U_x = V_y, U_y = -V_x$. The proof I have seen of this fact comes from Stein, and the key to the argument is to expand $U$ and $V$ via Taylor's formula for $C^1$ functions. That is, for a point $(x_0, y_0) \in \mathbb{R}^2$, we can write:

$$U(x,y) = U(x_0,y_0) + U_x(x_0,y_0)(x - x_0) + U_y(x_0, y_0)(y - y_0) + R(x,y),$$

and a similar formula for $V(x,y)$. Here, $R(x,y)$ is a remainder term with $\frac{R(x,y)}{|(x,y) - (x_0,y_0)|} \to 0$ as $(x,y) \to (x_0,y_0)$. I'm wondering if there is some way I can adapt this proof from Stein to the polar case?

Hints are solutions are greatly appreciated.

share|improve this question
2  
CR-equations in rectangular/polar coordinates are really the same equation except that they are in different coordinates. So the result for rectangular coordinates already imply that for polar coordinates. –  user27126 Jan 21 '13 at 2:22

1 Answer 1

up vote 1 down vote accepted

Here's the Taylor expansion approach. First, we need a linear algebra fact:

Any $\mathbb R$-linear map $T:\mathbb C\to\mathbb C$ can be written as $Tz=\alpha z+\beta\bar z$ for some (unique) $\alpha,\beta\in \mathbb C$.

Uniqueness should be clear, and existence follows by first writing $T(x+iy)=\gamma x+\delta y$ with complex $\gamma,\delta$ and then replacing $x=(z+\bar z)/2$, $y=(z-\bar z)/(2i)$.

This linear algebra fact is worth remembering, as it simplifies various computations in complex analysis.

Back to the problem. The first-order real Taylor expansion of $f$ at a point $z\in\mathbb C$ takes the form $$ f(z+h)=f(z)+\alpha h+\beta \bar h+o(|h|) $$ and we want to show that $\beta=0$ here. The equations you are given say that $r f_r + i f_\theta = 0$. Write $h=r e^{i\theta}$, and take the derivatives of $\alpha h+\beta \bar h$: $$ \begin{align} f_r &= \alpha e^{i\theta} + \beta e^{-i\theta} \\ f_\theta & = \alpha ir e^{i\theta} - \beta i r e^{-i\theta} \\ rf_r+f_\theta & = 2\beta r e^{-i\theta} \end{align} $$ Thus $\beta=0$ as desired.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.