I am working my way through Conway's Functions of One Complex Variable (2e), and I am a little unclear about some of the subtleties of complex conformal maps on the Riemann sphere versus in the complex plane.

For instance, on page 46, Conway defines a conformal map as "A function $f: G \rightarrow \mathbb{C}$ which has the angle preserving property and also has $$\lim_{z \to a} \frac{|f(z) - f(a)|}{|z-a|}$$

existing", where G is an open connected region of the complex plane. He goes on to say "If $f$ is analytic and $f'(z) \neq 0$ for any z, then $f$ is conformal.

My first question is, does every complex conformal map have to be analytic? As one example, consider Mobius transformations. For instance, $f(z) = 1/z$ is analytic with non-zero derivative everywhere in $\mathbb{C} \backslash \{0\}$, but it is not even defined at 0. However, since Mobius transformations are stated as being "exactly the bijective conformal maps from the Riemann sphere to itself", then obviously this map is conformal when viewed as a map on the Riemann sphere, but am I correct that technically it is not a conformal map on $\mathbb{C}$? Additionally, does this mean that $f(z) = 1/z$ is analytic on the extended complex plane $\widehat{\mathbb{C}}$? If so, can we always take a function with a single singularity in the complex plane and make it analytic on the extended complex plane?

I am a bit thrown off by the seemingly unstated transitions between when we are working in $\mathbb{C}$ versus in $\widehat{\mathbb{C}}$. As an example from one of the problems later in that section of Conway (p. 55, #14), he asks you to suppose one circle is contained inside another, with the circles tangent at the point a, and to find a conformal map from the region between the two circles onto the open unit disk using the hint (first try $(z-a)^{-1}$). However, this could only be conformal on the extended complex plane, correct?

I would greatly appreciate any help in sorting out exactly what subtleties are involved in determining when a complex function is conformal (and analytic as mentioned above) with the precise context of which area we are working (complex plane versus extended complex plane). Thanks!


Conformal implies holomorphic. Holomorphic functions are conformal at all points where their derivatives do not vanish.

  • 2
    $\begingroup$ But holomorphic is the same as complex analytic, so this still doesn't address the question of how the domain/codomain ($\mathbb{C}$ vs $\widehat{\mathbb{C}}$) affects the definition of analytic/holomorphic. If you read the full question, those subtleties are really the entire point of the question. $\endgroup$ Nov 12 '16 at 19:05

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