# Holomorphic function in poloar coordinates?

• Feb 9th 2010, 11:48 AM
davismj
Holomorphic function in poloar coordinates?
http://i45.tinypic.com/2utlrhw.jpg

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

$\displaystyle f(z) = f(\rho e^i\theta ) = u(\rho ,\theta ) + iv(\rho ,\theta )?$
• Feb 9th 2010, 03:22 PM
davismj
Quote:

Originally Posted by davismj
http://i45.tinypic.com/2utlrhw.jpg

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

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

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

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

However, taking the limit of $\displaystyle [f(x+h,y) - f(x,y)]/h$ in the purely real or imaginary sense still isn't getting me anywhere, or is it?
• Feb 9th 2010, 04:18 PM
dedust
Quote:

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

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

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

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

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

hence,

$\displaystyle r u_r = v_\theta$

$\displaystyle r v_r = -u_\theta$
• Feb 9th 2010, 04:25 PM
davismj
Quote:

Originally Posted by dedust
$\displaystyle \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} =\frac{u_{r}}{\sin \theta} + i \frac{v_{r}}{\sin \theta}$

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

hence,

$\displaystyle r u_r = v_\theta$

$\displaystyle r v_r = -u_\theta$

Thanks so much!