# Analytic Function Derivative

1. Jun 6, 2005

### sat

Take $x=r\cos\theta$ and $y=r\sin\theta$

If $f(z)=u(r,\theta) + iv(r,\theta)$, is analytic with u and v real, show that the derivative is given by
$$f'(z) = \left( \cos\theta \frac{\partial u}{\partial r}- \sin\theta\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( \cos\theta\frac{\partial v}{\partial r} - \sin\theta\frac{1}{r}\frac{\partial v}{\partial\theta} \right)$$

Since f is analytic, I use the result
$$f'(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$$
Though this seems to give
$$f'(z) = \left( (1/\cos\theta) \frac{\partial u}{\partial r}- (1/\sin\theta)\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left( (1/\cos\theta)\frac{\partial v}{\partial r} - (1/\sin\theta)\frac{1}{r}\frac{\partial v}{\partial\theta} \right)$$

Can anyone see why this isn't correct?

2. Jun 6, 2005

### dextercioby

How did you apply the chain rule...?I think there's a problem with that.

Daniel.

3. Jun 6, 2005

### sat

I just did:

du/dx = (du/dr)(dr/dx) + (du/dtheta)(dtheta/dx)

though I think that's probably not quite right when I think about the dependence of the variables on eachother...

Though the "units" look OK at first sight, I'm not sure that I can do that.

4. Jun 6, 2005

### dextercioby

Here's where you went wrong

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial\theta}\frac{\partial\theta}{\partial x}$$ (1)

$$\frac{\partial r}{\partial x}=\frac{\partial}{\partial x}\sqrt{x^{2}+y^{2}}=\frac{x}{\sqrt{x^{2}+y^{2}}}=\frac{r\cos\theta}{r}=\cos\theta$$ (2)

Can u do the other derivative...?

Daniel.

5. Jun 6, 2005

### sat

Thanks. I'll look at that.