Let \(f(z)=z+1 / z\). Describe the images of both the circle \(|z|=r\) of radius \(r(r \neq 0)\) and the ray \(\arg z=\theta_{0}\) under \(f\) in terms of well known curves.
Solution
Write \(z=re^{i\theta}\).
For the circle \(|z|=r\), \[ w=z+\frac1z =\left(r+\frac1r\right)\cos\theta +i\left(r-\frac1r\right)\sin\theta. \] Thus, for \(r\ne1\), its image is the ellipse \[ \boxed{ \frac{(\Re w)^2}{(r+r^{-1})^2} +\frac{(\Im w)^2}{(r-r^{-1})^2}=1.} \] Its foci are at \(\pm2\). When \(r=1\), the ellipse degenerates to the segment \([-2,2]\) on the real axis.
For the ray \(\arg z=\theta_0\), write \(z=\rho e^{i\theta_0}\) with \(\rho>0\). Then \[ x=(\rho+\rho^{-1})\cos\theta_0, \qquad y=(\rho-\rho^{-1})\sin\theta_0. \] Since \[ (\rho+\rho^{-1})^2-(\rho-\rho^{-1})^2=4, \] we obtain, when both sine and cosine are nonzero, \[ \boxed{ \frac{x^2}{4\cos^2\theta_0} -\frac{y^2}{4\sin^2\theta_0}=1.} \] Hence the image is one branch of a hyperbola with foci at \(\pm2\). For \(\sin\theta_0=0\) it degenerates to a real ray outside \([-2,2]\); for \(\cos\theta_0=0\) it degenerates to the imaginary axis.