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?

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    $\begingroup$ How can the function take values in $\mathbb{R}^2$ if $\lambda$ is complex? $\endgroup$
    – Eric Auld
    Jul 23, 2013 at 10:23
  • $\begingroup$ Yes, sorry, that indeed is a problem. Actually I made a mistake there, I'll fix it. Thanks! $\endgroup$ Jul 23, 2013 at 10:32
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    $\begingroup$ Isn't the function simply $id_{\mathbb{C}}^{\lambda}$? $\endgroup$
    – scineram
    Jul 23, 2013 at 10:50
  • $\begingroup$ I'm sorry, I'm not familiar with this notation. $\endgroup$ Jul 23, 2013 at 10:55
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    $\begingroup$ 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 $\lambda = 1/2$. $\endgroup$ Jul 23, 2013 at 11:44

1 Answer 1


First, one should be aware that the word "conformal" is used in different ways. Often it's "holomorphic and bijective". Other times, holomorphic with nonzero derivative, as here.

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 $\lambda\in\mathbb Z$.


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