1

What is wrong with my derivation here (I am missing a factor of 1/2, but I'm not sure where from):

Let $f(z)= u(r,\theta)+iv(r, \theta)$ and $z=re^{i\theta}$.

Then, $$\frac{df}{dz}=\frac{du}{dz}+i\frac{dv}{dz}$$ $$=\frac{\partial u}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial z}+i(\frac{\partial v}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial v}{\partial \theta}\frac{\partial \theta}{\partial z})$$

Using $z=re^{i \theta}$, we have $\frac{\partial z}{\partial r}= e^{i\theta}$ and $\frac{\partial z}{\partial \theta}= ire^{i\theta}$ so that $\frac{\partial r}{\partial z}= e^{-i\theta}$ and $\frac{\partial \theta}{\partial z}= \frac{-ie^{-i\theta}}{r}$.

Substituting these in and using the Cauchy-Riemann conditions in polar form we have: $$=\frac{\partial u}{\partial r}e^{-i\theta}+\frac{\partial u}{\partial \theta}\frac{-ie^{-i\theta}}{r}+i(\frac{\partial v}{\partial r}e^{-i\theta}+\frac{\partial v}{\partial \theta}\frac{-ie^{-i\theta}}{r})$$ $$=\frac{\partial u}{\partial r}e^{-i\theta}-r\frac{\partial v}{\partial r}\frac{-ie^{-i\theta}}{r}+i(\frac{\partial v}{\partial r}e^{-i\theta}+r\frac{\partial u}{\partial r}\frac{-ie^{-i\theta}}{r})$$ $$=2(\frac{\partial u}{\partial r}e^{-i\theta}+i(\frac{\partial v}{\partial r}e^{-i\theta}))$$

Which is off by a factor of 2. I believe in flipping partial derivatives, I have made a poor assumption, but I am not sure where. Any help would be appreciated.

Thomas Clark
  • 69

1 Answers1

3

The error

It is true that $\dfrac{\partial z}{\partial r}=e^{i\theta}$ and $\dfrac{\partial z}{\partial\theta}=ire^{i\theta}$.

But even in $\mathbb{R}^{2}\to\mathbb{R}^{2}$, you cannot simply take a reciprocal to flip a partial derivative. See Mark Viola's answer to "Partial derivatives inverse question".

And with $\mathbb{C}$, you can't simply apply the matrix version of the chain rule as in that answer, because you don't immediately have a $2\times2$ matrix with one column corresponding to $\dfrac{\partial}{\partial z}$. To make your approach work out, you'd need to have a version of $\dfrac{\partial}{\partial z}$ that works for functions that aren't complex-differentiable and pair $\dfrac{\partial}{\partial z}$ with something else to make a matrix. And since you started with $\dfrac{\mathrm{d}}{\mathrm{d}z}$ and changed it to $\dfrac{\partial}{\partial z}$, you'd need to pair it with something new that is $0$ when a function is complex-differentiable.

Fixing your approach

It turns out that a nice pair exists in the Wirtinger derivatives $\dfrac{\partial}{\partial z}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)$ and $\dfrac{\partial}{\partial\overline{z}}=\dfrac{1}{2}\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)$, which satisfy things like $\mathrm{d}f=\dfrac{\partial f}{\partial z}\mathrm{d}\left(x+iy\right)+\dfrac{\partial f}{\partial\overline{z}}\mathrm{d}\left(x-iy\right)$.

Then we have (speculatively, at least): $$\begin{bmatrix}\dfrac{\partial z}{\partial r} & \dfrac{\partial z}{\partial\theta}\\ \dfrac{\partial\overline{z}}{\partial r} & \dfrac{\partial\overline{z}}{\partial\theta} \end{bmatrix}\begin{bmatrix}\dfrac{\partial r}{\partial z} & \dfrac{\partial r}{\partial\overline{z}}\\ \dfrac{\partial\theta}{\partial z} & \dfrac{\partial\theta}{\partial\overline{z}} \end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}$$

To use this, we first calculate $\dfrac{\partial\overline{z}}{\partial r}=\cos\theta-i\sin\theta=e^{-i\theta}$ and $\dfrac{\partial\overline{z}}{\partial\theta}=-r\sin\theta-ir\cos\theta=r\sin\left(-\theta\right)-ir\cos\left(-\theta\right)$ $=-\left(-r\sin\left(-\theta\right)+ir\cos\left(-\theta\right)\right)=-ire^{-i\theta}$.

$$\begin{bmatrix}e^{i\theta} & ire^{i\theta}\\ e^{-i\theta} & -ire^{-i\theta} \end{bmatrix}\begin{bmatrix}\dfrac{\partial r}{\partial z} & \dfrac{\partial r}{\partial\overline{z}}\\ \dfrac{\partial\theta}{\partial z} & \dfrac{\partial\theta}{\partial\overline{z}} \end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}$$

$$\begin{bmatrix}\dfrac{\partial r}{\partial z} & \dfrac{\partial r}{\partial\overline{z}}\\ \dfrac{\partial\theta}{\partial z} & \dfrac{\partial\theta}{\partial\overline{z}} \end{bmatrix}=\dfrac{1}{e^{i\theta}\left(-ire^{-i\theta}\right)-ire^{i\theta}e^{-i\theta}}\begin{bmatrix}-ire^{-i\theta} & -ire^{i\theta}\\ -e^{-i\theta} & e^{i\theta} \end{bmatrix}$$

$$\begin{bmatrix}\dfrac{\partial r}{\partial z} & \dfrac{\partial r}{\partial\overline{z}}\\ \dfrac{\partial\theta}{\partial z} & \dfrac{\partial\theta}{\partial\overline{z}} \end{bmatrix}=\dfrac{i}{2r}\begin{bmatrix}-ire^{-i\theta} & -ire^{i\theta}\\ -e^{-i\theta} & e^{i\theta} \end{bmatrix}$$

$$\begin{bmatrix}\dfrac{\partial r}{\partial z} & \dfrac{\partial r}{\partial\overline{z}}\\ \dfrac{\partial\theta}{\partial z} & \dfrac{\partial\theta}{\partial\overline{z}} \end{bmatrix}=\begin{bmatrix}\dfrac{1}{2}e^{-i\theta} & \dfrac{1}{2}e^{i\theta}\\ \dfrac{-ie^{-i\theta}}{2r} & \dfrac{ie^{i\theta}}{2r} \end{bmatrix}$$

Thus, it seems like $\dfrac{\partial r}{\partial z}=\dfrac{1}{2}e^{-i\theta}$ and $\dfrac{\partial\theta}{\partial z}=\dfrac{1}{2}\dfrac{-ie^{-i\theta}}{r}$, which fixes the factor of $2$ in your calculations.

Another approach

Developing/justifying/using the Wirtinger derivatives is a lot of work just for this problem, so here is a simpler approach to the original problem. Note/recall that the complex derivative equals the derivative taken along any direction (e.g. the $x$ direction), so that we have: $$\dfrac{\mathrm{d}f}{\mathrm{d}z}=\dfrac{\partial f}{\partial x}=\dfrac{\partial u}{\partial x}+i\dfrac{\partial v}{\partial x}$$ $$=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial\theta}\dfrac{\partial\theta}{\partial x}+i\left(\dfrac{\partial v}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial v}{\partial\theta}\dfrac{\partial\theta}{\partial x}\right)$$

Now, $r=\sqrt{x^{2}+y^{2}}$, so $\dfrac{\partial r}{\partial x}=\dfrac{x}{\sqrt{x^{2}+y^{2}}}=\dfrac{x}{r}$. And for most points in the plane, we have $\theta=\begin{cases} \arctan\left(\dfrac{y}{x}\right) & \text{ if }x>0\\ \arctan\left(\dfrac{y}{x}\right)+\pi & \text{ if }x>0\text{ and }y\ge0\\ \arctan\left(\dfrac{y}{x}\right)-\pi & \text{ if }x,y<0 \end{cases}$, so $\dfrac{\partial\theta}{\partial x}=\dfrac{-y/x^{2}}{\left(y/x\right)^{2}+1}=-\dfrac{y}{x^{2}+y^{2}}=-\dfrac{y}{r^{2}}$.

This gives us:

$$\dfrac{\mathrm{d}f}{\mathrm{d}z}=\dfrac{\partial u}{\partial r}\dfrac{x}{r}-\dfrac{\partial u}{\partial\theta}\dfrac{y}{r^{2}}+i\left(\dfrac{\partial v}{\partial r}\dfrac{x}{r}-\dfrac{\partial v}{\partial\theta}\dfrac{y}{r^{2}}\right)$$

And the Cauchy-Riemann equations yield:

$$=\dfrac{\partial u}{\partial r}\dfrac{x}{r}+r\dfrac{\partial v}{\partial r}\dfrac{y}{r^{2}}+i\left(\dfrac{\partial v}{\partial r}\dfrac{x}{r}-r\dfrac{\partial u}{\partial r}\dfrac{y}{r^{2}}\right)$$

$$=\dfrac{\partial u}{\partial r}\dfrac{x-iy}{r}+\dfrac{\partial v}{\partial r}\dfrac{ix+y}{r}$$

$$=\dfrac{x-iy}{r}\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)$$

$$=e^{-\theta}\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)$$

Mark S.
  • 23,925