<s>[INST] Complex transformation $ w = z + \frac{c}{z} $ where $ |z| = 1 $ 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.
 [/INST] $z$ 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*}
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