# Show that map is conformal

I want to show that the map $\phi(r,\theta) = r^\lambda (\cos(\lambda \theta), \sin(\lambda \theta))$, where $\lambda \in \mathbb{C}$, is conformal on the slit plane $\{(r,\theta)| r > 0, -\pi < \theta < \pi \}$. $(r,\theta)$ are the standard polar coordinates in $\mathbb{R}^2 = \mathbb{C}$

What I did is i wrote $\phi(re^{i\theta}) = r^\lambda e^{i\lambda \theta}$. I defined $r^\lambda = e^{\lambda \log r}$. Then I verified with Cauchy-Riemann in polar coordinates, that the function is holomorphic, and that the derivative has no zeros. If I'm correct, this suffices. (I found that I need $\lambda \neq 0$).

Next, I wanted to find out for which $\lambda \in \mathbb{C}$ the function is conformal on $\mathbb{R}^2 \setminus \{0\}$. However, I found no restriction, as the logarithm I need to define the power doesn't need to be complex, as it only takes real, positiv arguments. Am I right? Or am I missing something?

• How can the function take values in $\mathbb{R}^2$ if $\lambda$ is complex? Jul 23, 2013 at 10:23
• Yes, sorry, that indeed is a problem. Actually I made a mistake there, I'll fix it. Thanks! Jul 23, 2013 at 10:32
• Isn't the function simply $id_{\mathbb{C}}^{\lambda}$? Jul 23, 2013 at 10:50
• I'm sorry, I'm not familiar with this notation. Jul 23, 2013 at 10:55
• In order for $\phi$ to be conformal on the punctured plane $\mathbf{C} \setminus \{0\}$, it must first be defined everywhere on the punctured plane. A straightforward approach is separately to define $\phi$ on a "copy" of the complex plane slit along the positive real axis, and to check whether your two definitions agree off the real axis. If that doesn't help, try a specific value, such as $\lambda = 1/2$. Jul 23, 2013 at 11:44

On the slit plane, we can use the principal value of $\log z$ to define $z^\lambda= \exp(\lambda \log z)$ which makes $z^\lambda$ a holomorphic function for all $\lambda\in\mathbb C$. The derivative does not vanish if $\lambda\ne 0$.
In order to extend the above function holomorphically to the punctured plane, we need the limits of $z^\lambda$ to be the same when $z=re^{i\theta}$, with $\theta\to \pm\pi$. This requires $\exp(2\pi i\lambda)=1$, which happens only when $\lambda\in\mathbb Z$.