Answer to a query concerning the mapping \(w=z^{1/m}\) (Q1319098)

This is an attractive, well-written computation using elementary differential geometry. Let \(D= \{| z-1| <r\}\) for \(0<r<1\) and consider the function \(f(z)= z^{1/m}\) on \(D\), \(m\geq 1\), \(m\) real. Let \(C_{r,m}\) be the disk with the images of \((1\pm r)\) under \(f\) being endpoints of the diameter; call these points \(\alpha\) and \(\beta\). Then \(f(D)\subset C_{r,m}\). This answers a question of \textit{L. Petkovíć} [Query 359, Notices, Am. Math. Soc. 33, No. 4, 629 (1986)] in the affirmative. The proof is to observe that \(\partial C\) is tangent to \(\partial D\) at \(\alpha\) and \(\beta\), and then compute the curvature \(\kappa\) of the curve \(w(\theta)= (1+re^{i\theta} )^ m\), \(0\leq \theta\leq 2\pi\). The fact that \(\kappa (\theta)\) has four simple zeros on \([0, 2\pi)\) plays an essential role.