Sign up ×
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.

How would one go about showing the polar version of the Cauchy Riemann Equations are sufficient to get differentiability of a complex valued function which has continuous partial derivatives?

I haven't found any proof of this online.

One of my ideas was writing out r and theta in terms of x and y, then taking the partial derivatives with respect to x and y and showing the Cauchy Riemann equations in the Cartesian coordinate system are satisfied. A problem with this approach is that derivatives get messy.

What are some other ways to do it?

share|cite|improve this question

5 Answers 5

up vote 11 down vote accepted

I happen to have some notes on this question. What follows here is the usual approach, it's just multivariate calculus paired with the Cauchy Riemann equations. I have an idea for an easier way, I'll post it as a second answer in a bit if it works.

If we use polar coordinates to rewrite $f$ as follows: $$ f(x(r,\theta),y(r,\theta)) = u(x(r,\theta),y(r,\theta))+iv(x(r,\theta),y(r,\theta)) $$ we use shorthands $F(r,\theta)=f(x(r,\theta),y(r,\theta))$ and $U(r,\theta )=u(x(r,\theta),y(r,\theta))$ and $V(r,\theta )=v(x(r,\theta),y(r,\theta))$. We derive the CR-equations in polar coordinates via the chain rule from multivariate calculus, $$ U_r = x_ru_x + y_ru_y = \cos(\theta)u_x + \sin(\theta)u_y \ \ \text{and} \ \ U_{\theta} = x_{\theta}u_x + y_{\theta}u_y = -r\sin(\theta)u_x + r\cos(\theta)u_y $$ Likewise, $$ V_r = x_rv_x + y_rv_y = \cos(\theta)v_x + \sin(\theta)v_y \ \ \text{and} \ \ V_{\theta} = x_{\theta}v_x + y_{\theta}v_y = -r\sin(\theta)v_x + r\cos(\theta)v_y $$ We can write these in matrix notation as follows: $$ \left[ \begin{array}{l} U_r \\ U_{\theta} \end{array} \right] = \left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]\left[ \begin{array}{l} u_x \\ u_y \end{array} \right] \ \ \text{and} \ \ \left[ \begin{array}{l} V_r \\ V_{\theta} \end{array} \right] = \left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]\left[ \begin{array}{l} v_x \\ v_y \end{array} \right] $$ Multiply these by the inverse matrix: $\left[ \begin{array}{ll} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{array} \right]^{-1} = \frac{1}{r}\left[ \begin{array}{ll} r\cos(\theta) & -\sin(\theta) \\ r\sin(\theta) & \cos(\theta) \end{array} \right]$ to find $$ \left[ \begin{array}{l} u_x \\ u_y \end{array} \right] = \frac{1}{r}\left[ \begin{array}{ll} r\cos(\theta) & -\sin(\theta) \\ r\sin(\theta) & \cos(\theta) \end{array} \right]\left[ \begin{array}{l} U_r \\ U_{\theta} \end{array} \right] = \left[ \begin{array}{l} \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} \\ \sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} \end{array} \right] $$ A similar calculation holds for $V$. To summarize: $$ u_x = \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} \ \ \ \ v_x = \cos(\theta)V_r - \tfrac{1}{r}\sin(\theta)V_{\theta} $$ $$ u_y =\sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} \ \ \ \ v_y =\sin(\theta)V_r + \tfrac{1}{r}\cos(\theta)V_{\theta} $$ The CR-equation $u_x=v_y$ yields: $$ (A.) \ \ \cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta} = \sin(\theta)V_r + \tfrac{1}{r}\cos(\theta)V_{\theta} $$ Likewise the CR-equation $u_y=-v_x$ yields: $$ (B.) \ \ \sin(\theta)U_r + \tfrac{1}{r}\cos(\theta)U_{\theta} = -\cos(\theta)V_r + \tfrac{1}{r}\sin(\theta)V_{\theta}$$ Multiply (A.) by $r\sin(\theta)$ and $(B.)$ by $r\cos(\theta)$ and subtract (A.) from (B.): $$ \boxed{U_{\theta} = -rV_r} $$ Likewise multiply (A.) by $r\cos(\theta)$ and $(B.)$ by $r\sin(\theta)$ and add (A.) and (B.): $$ \boxed{rU_r = V_{\theta}} $$ Finally, recall that $z = re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$ hence \begin{align} \notag f'(z) &= u_x+iv_x \\ \notag &= (\cos(\theta)U_r - \tfrac{1}{r}\sin(\theta)U_{\theta})+i(\cos(\theta)V_r - \tfrac{1}{r}\sin(\theta)V_{\theta}) \\ \notag &= (\cos(\theta)U_r + \sin(\theta)V_{r})+i(\cos(\theta)V_r - \sin(\theta)U_{r}) \\ \notag &= (\cos(\theta)- i\sin(\theta))U_r + i(\cos(\theta)-i\sin(\theta))V_r \\ \notag &= e^{-i\theta}( U_r+iV_r) \notag \end{align}

share|cite|improve this answer
I see, I think the only point of possible problem is knowing that x and y do show up as part of the function, – Mike Oct 2 '12 at 1:23
@Mike well, we are taking a fact known for $x,y$ and trying to change notation to polar coordinates. There will be $x$ and $y$ somewhere. The key is understanding how to convert both functions and derivatives. You can see if you find my other answer reasonable... there is a jump, see if you can find it. – James S. Cook Oct 2 '12 at 1:40
Can we express the CR equations just by using equation A, namely equating the terms adjacent to sine and cosine on both sides? – Shemafied Oct 27 at 8:49
@Shemafied I suppose that would also be reasonable. Interesting. – James S. Cook Oct 28 at 0:16
@james Yes, but the proof you showed is mathematically more sound – Shemafied Oct 29 at 10:14

A less standard approach: Take $\{e_{r},e_{\theta}\}$ as a basis in polar coordinates then the Jacobian matrix for any function on $\mathbb{R}^2$ has the form: $$ [df] = \left[ \begin{array}{cc} U_r & U_{\theta} \\ V_r & V_{\theta} \end{array} \right] $$ Define complex-differentiability via complex linearity of the differential; $df(vw)=df(v)w$ at the point in question for all $v,w \in \mathbb{C}$. In particular this gives the beautiful formula: $df(v)=df(1)v$; this means the first column of the Jacobian fixes the second by complex multiplication. It is geometrically clear that $\frac{1}{r}e_{\theta} = ie_r$. Observe that $e_{\theta} = ire_r$ thus $df(e_{\theta}) =df(ire_{r})=irdf(e_r)$. On the other hand, $f=U+iV$ and $$ df(e_r) = U_r+iV_r \qquad \& \qquad df(e_{\theta})=f_{\theta}=U_{\theta}+iV_{\theta} $$ Thus, $U_{\theta}+iV_{\theta}=ir(U_r+iV_r)$ and we derive $$ \boxed{ U_{\theta} = -rV_r \qquad \& \qquad V_{\theta}=rU_r. } $$

share|cite|improve this answer
I would really love to be able to follow through your answer, but I shold first freshen up my Linear arlebra – Shemafied Oct 27 at 23:44

We can derive using purely polar coordinates. Start with \begin{align} z(r,\theta) &=r\,\mathrm{e}^{\mathrm{i}\theta} \\ f(z) &= u(r,\theta) + \mathrm{i} v(r,\theta) \end{align} We define $f'(z)$ using the limit $$ f'(z) = \lim_{z\to 0} \frac{\Delta f}{\Delta z} $$ where \begin{align} \Delta f &= \Delta u + \mathrm{i} \Delta v \\ \Delta z &= z(r+\Delta r,\theta + \Delta\theta) - z(r,\theta) \\ &= (r+\Delta r)\,\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)} - r \,\mathrm{e}^{\mathrm{i}\theta} \end{align} Next, we try to first approach from $\Delta\theta\to 0$, $$ \Delta z = (r+\Delta r)\,\mathrm{e}^{\mathrm{i}\theta} - r \,\mathrm{e}^{\mathrm{i}\theta} = \Delta r\,\mathrm{e}^{\mathrm{i}\theta} $$ Therefore when we take the limit, $$ f'(z) = \lim_{\Delta r\to 0} \frac{\Delta u + \mathrm{i} \Delta v}{\Delta r \,\mathrm{e}^{i\theta}} = \frac{1}{\mathrm{e}^{\mathrm{i}\theta}}(u_r + \mathrm{i} v_r) \label{a}\tag{1}$$ On the other hand, approaching from $\Delta r\to 0$ first yields $$\Delta z = r\,\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)} - r \,\mathrm{e}^{\mathrm{i}\theta}$$ Using the derivative $$\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)}-\mathrm{e}^{\mathrm{i}\theta} = \frac{\mathrm{d}\mathrm{e}^{\mathrm{i}\theta}}{\mathrm{d}\theta}\Delta\theta = \mathrm{i}\mathrm{e}^{\mathrm{i}\theta}\Delta\theta$$ we get $$\Delta z = \mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}\Delta\theta$$ Therefore $$f'(z) = \lim_{\Delta\theta\to 0} \frac{\Delta u + \mathrm{i} \Delta v}{\mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}\Delta\theta} = \frac{1}{\mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}}(u_\theta + \mathrm{i}v_\theta) \label{b}\tag{2}$$ Finally, comparing the real and imaginary parts of ($\ref{a}$) and ($\ref{b}$) gives us what we want: $$u_r=\frac{1}{r}v_\theta \quad,\quad v_r = -\frac{1}{r}u_\theta$$

Reference: Kwong-tin Tang, Mathematical Methods for Engineers and Scientists 1 (Springer, 2007)

share|cite|improve this answer
I think this is a standard approach though at its present form it is not completely rigorous. But surely this can be made rigorous by applying some MVT argument along one of the lines above. – julypraise Dec 23 '14 at 19:29

Here is an easy approach Here is an easy approach to derive the Cauchy-Riemann equations in polar form.

share|cite|improve this answer

I understand this topic is somewhat old, but I still feel like I can contribute to it with a derivation in my eyes much simpler.

The key is to understand the goal: we want to derive some analogue of the CR-equations, except in polar form. We therefore wish to relate $u_\theta$ with $v_r$ and $v_\theta$ with $u_r$.

Take the CR-equations in the $xy$-plane as an example and expand: \begin{equation} u_x = v_y \Leftrightarrow u_\theta \theta_x = v_r r_y. \end{equation} Now remember the definitions of polar coordinates and take the appropriate derivatives: \begin{align*} x = r\cos\theta \Rightarrow x_\theta = -r\sin\theta \Rightarrow \theta_x = \frac1{-r\sin\theta}\\ y = r\sin\theta \Rightarrow y_r = \sin\theta \Rightarrow r_y = \frac1{\sin\theta}. \end{align*} Now fill in the expressions into our expanded CR-equation to find \begin{align*} u_\theta \frac1{-r\sin\theta} = v_r \frac1{\sin\theta} \Rightarrow u_\theta = -r v_r. \end{align*} In a similar fashion we can also derive \begin{equation} u_r = \frac1r v_\theta. \end{equation}

share|cite|improve this answer

Your Answer


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.