# Confusion about partial differentiation and Cauchy-Riemann equations

In complex analysis class (using Stein's Complex Analysis), we learned about the derivation of the Cauchy-Riemann equations, and that made sense. We take a holomorphic $f : O \rightarrow \mathbb{C}$, where $O \subset \mathbb{C}$ is open, and split it as $f(x + iy) = u(x, y) + i v(x, y)$. Then after some computation we arrive at $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$. However, I haven't learned about multivariable calculus so I'm new to partial differentiation. The above makes sense to me, but in an exercise they ask us to prove that, in polar form, these equations take the form $\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$ and $\frac{1}{r} \frac{\partial u}{\partial \theta} = - \frac{\partial v}{\partial r}$. Now I'm confused. Do they mean the same $u$ and $v$ we had been using before? Maybe they similarly define $f(re^{i\theta}) = u(r, \theta) e^{i v(r, \theta)}$ and want me to derive the equations for that? The first option doesn't make any sense to me and I didn't succeed at attempting the second. I think I need a pretty thorough clarification on this issue, can anyone help?

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## 2 Answers

Since you're not familiar with multivariable calc, let me try to clarify:

If we consider points in $\mathbb{R}^2$ using polar coordinates, we have, as anon said, $x=r\cos\theta$ and $y=r\sin\theta$. This means $$u(x,y)=u(x(r,\theta),y(r,\theta))=u(r\cos\theta,r\sin\theta).$$ Using the chain rule, we get $$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}(x(r,\theta),y(r,\theta))\cdot\frac{\partial x}{\partial r}(r,\theta)+\frac{\partial u}{\partial y}(x(r,\theta),y(r,\theta))\cdot\frac{\partial y}{\partial r}(r,\theta)$$ and $$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}(x(r,\theta),y(r,\theta))\cdot\frac{\partial x}{\partial \theta}(r,\theta)+\frac{\partial u}{\partial y}(x(r,\theta),y(r,\theta))\cdot\frac{\partial y}{\partial \theta}(r,\theta).$$

As we can find $\frac{\partial x}{\partial r}, \frac{\partial y}{\partial r}, \frac{\partial x}{\partial \theta}$ and $\frac{\partial y}{\partial \theta}$ explicitly in terms of $r$ and $\theta$, then we get a system of equations that lets us solve for $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ in terms of $r$, $\theta$, $\frac{\partial u}{\partial r}$, and $\frac{\partial u}{\partial \theta}$. From here it is just a matter of plugging the alternate expressions for $u_x$ and $u_y$ back into the Cauchy-Riemann equations that you know and love and simplifying.

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No, $u$ and $v$ are still the real and imaginary parts of $f$, but $\mathbb{C}$ (as the domain, not codomain) is now taken to be in polar form. Explicitly, we have $x=r\cos\theta$ and $y=r\sin\theta$; substitute these expressions into $u(x,y)$ and $v(x,y)$, then perform partial differentiation with respect to $r$ and $\theta$ with chain rule.

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