Holomorphic function in poloar coordinates?

**davismj** · February 9th 2010, 11:48 AM

Okay, so if $f(z) = u(x,y) + iv(x,y)$ , then in polar coordinates, since $z = \rho e^i\theta$ , then...

$f(z) = f(\rho e^i\theta ) = u(\rho ,\theta ) + iv(\rho ,\theta )?$

**davismj** · February 9th 2010, 03:22 PM

Originally Posted by davismj

Okay, so if $f(z) = u(x,y) + iv(x,y)$ , then in polar coordinates, since $z = \rho e^i\theta$ , then...

$f(z) = f(\rho e^i\theta ) = u(\rho ,\theta ) + iv(\rho ,\theta )?$

After thinking about it, I realized that while the functions u and v would be dependent on $\rho$ and $\theta$ , the real and imaginary parts would both be dependent on $\rho$ and $\theta$ , and therefore it's probably more accurate to represent them thus:

$f(z) = f(e^i\theta ) = u(\rho cos \theta ,\rho sin \theta) + iv(\rho cos \theta ,\rho sin \theta)$

However, taking the limit of $[f(x+h,y) - f(x,y)]/h$ in the purely real or imaginary sense still isn't getting me anywhere, or is it?

**dedust** · February 9th 2010, 04:18 PM

Originally Posted by davismj

After thinking about it, I realized that while the functions u and v would be dependent on $\rho$ and $\theta$ , the real and imaginary parts would both be dependent on $\rho$ and $\theta$ , and therefore it's probably more accurate to represent them thus:

$f(z) = f(e^i\theta ) = u(\rho cos \theta ,\rho sin \theta) + iv(\rho cos \theta ,\rho sin \theta)$

However, taking the limit of $[f(x+h,y) - f(x,y)]/h$ in the purely real or imaginary sense still isn't getting me anywhere, or is it?

$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} =\frac{u_{r}}{\sin \theta} + i \frac{v_{r}}{\sin \theta}$

$i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} = - i \frac{u_{\theta}}{r \sin \theta} + \frac{v_{\theta}}{r \sin \theta}$

hence,

$r u_r = v_\theta$

$r v_r = -u_\theta$

**davismj** · February 9th 2010, 04:25 PM

Originally Posted by dedust

$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} =\frac{u_{r}}{\sin \theta} + i \frac{v_{r}}{\sin \theta}$

$i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} = - i \frac{u_{\theta}}{r \sin \theta} + \frac{v_{\theta}}{r \sin \theta}$

hence,

$r u_r = v_\theta$

$r v_r = -u_\theta$

Thanks so much!

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