Abstract. We answer a question raised by M. Chuaqui, P. Duren, and B. Osgood by showing that a conformal mapping of a simply connected domain cannot take two circles onto two proper ellipses.The goal of this note is to prove the following.
Proposition 1.Let Ω be a simply connected domain on C = C ∪ {∞} containing circles (on C) C 1 and C 2 , C 1 = C 2 . Let f be a conformal mapping from Ω into C. If f maps each of the circles C 1 and C 2 onto an ellipse, then f is a Möbius transformation and therefore the ellipses are actually circles.Thus this proposition answers affirmatively a question raised by M. Chuaqui, P. Duren, and B. Osgood in [1] that a conformal mapping of a simply connected domain cannot take two circles onto two ellipses.Precomposing f with a Möbius transformation and postcomposing it with a linear transformation, we may assume that C 1 = {z : |z| = 1} and that f maps the unit disc D = {z : |z| < 1} onto the interior of an ellipse L 1 with foci ±1 such that f (r) = 1, f (−r) = −1 for some 0 < r < 1. Then it is a standard exercise in complex analysis to verify that w = f (z) is defined bywhere K(·) denotes the complete elliptic integral of the first kind. Since the image f (D) is symmetric with respect to the coordinate axes, it follows that f is an odd function having real coefficients, i.e.,We want to stress that our proof below does not require the explicit form (1) of the mapping function, which is included here for completeness. Instead, we will use the fact that every ellipse L having foci w 1 and w 2 is a trajectory of the quadratic differential (2) Q 1 (w) dw 2 = − dw 2 (w − w 1 )(w − w 2 ) ;