Complex transformation $ w = z + \frac{c}{z} $ where $ |z| = 1 $

Question

I am to show that if $ w = z + \frac{c}{z} $ and $ |z| = 1 $, then $w$ is an ellipse, and I must find its equation.

Previously, I have solved transformation questions by finding the modulus of the transformation in either the form $ w = f(z) $ or $ z = f(w) $. However, I think the part stumping me here is that the transformation has both $z$ and $z^{-1}$ in it.

Attempt with $w = f(z)$: $$ w = x+iy + \frac{c}{x+iy} $$ $$= x+iy + \frac{c(x-iy)}{x^2 + y^2} $$ $$= x+iy+c(x-iy)$$ [as $|z| = 1 \implies x^2 + y^2 = 1^2$]

However, from here I am unable to work out how to proceed.

I also tried to find $z=f(w)$: $$ z^2 - zw + c = 0 $$ $$ (z - \frac{w}{2})^2 + c - \frac{w^2}{4} = 0 $$ But I cannot see how this is of any use either.

Answer 1

$z$ is on the unit circle; let $z=e^{i \theta}$ so \begin{eqnarray*} w= (1+c) \cos( \theta) +i (1-c) \sin(\theta) \end{eqnarray*} which gives $ x= (1+c) \cos( \theta) , y= (1-c) \sin(\theta)$ and considered in cartesian coordinates \begin{eqnarray*} \frac{x^2}{ (1+c)^2} + \frac{y^2}{ (1-c)^2} =1 . \end{eqnarray*}

Answer 2

Let $\,z=u^2\,$ with $\,|u| = \sqrt{|z|}=1\,$, and let $\,c=b^2\,$. Then $\displaystyle\,w=u^2+\frac{b^2}{u^2}\,$, and so:

$$ w+2b=u^2+\frac{b^2}{u^2}+ 2b=\left(u+\frac{b}{u}\right)^2 \\ \implies |w+2b|=\left|u+\frac{b}{u}\right|^2 = \left(u+\frac{b}{u}\right)\left(\bar u+\frac{\bar b}{\bar u}\right) = |u|^2+\frac{|b|^2}{|u|^2}+\frac{b \bar u}{u}+\frac{\bar b u}{\bar u} \\[5px] w-2b=u^2+\frac{b^2}{u^2}- 2b=\left(u-\frac{b}{u}\right)^2 \\ \implies |w-2b|=\left|u-\frac{b}{u}\right|^2 = \left(u-\frac{b}{u}\right)\left(\bar u-\frac{\bar b}{\bar u}\right) = |u|^2+\frac{|b|^2}{|u|^2}-\frac{b \bar u}{u}-\frac{\bar b u}{\bar u} $$

Adding the above gives:

$$ |w+2b| + |w-2b| = 2|u|^2+2\,\frac{|b|^2}{|u|^2} = 2(1+|b|^2) \tag{1} $$

Therefore the locus of $\,w\,$ is an ellipse with foci $\,\pm 2b\,$ and semi-major axis $1+|b|^2 \ge 2|b|\;$ (which degenerates to the segment $\,[-2b,2b]\,$ iff $\,|b|=1 \iff |c|=1\,$).

---

[ EDIT ] To elaborate on $(1)$, the equation says that the sum of distances from $w$ to the fixed points $-2b$ and $2b$ is constant, which is the geometric definition of an ellipse. In general, the equation $|w-a|+|w-b|=c \in \mathbb{R}^+$ describes an ellipse with foci $a,b \in \mathbb{C}$ and semi-major axis $\frac{1}{2}c$ when $c \gt |a-b|$. The locus degenerates to the segment $[a,b]$ if $c=|a-b|$ and is empty if $c \lt |a-b|$.