Locus $\{e^{i \theta} + be^{-i \theta} \ : \theta \in [0, 2 \pi] \}$ for fixed $b \in \mathbb{C}$ 
What is the locus of $\{e^{i \theta} + be^{-i \theta} \ : \theta \in [0, 2 \pi]  \}$ for fixed $b \in \mathbb{C}$?

Write $ w := u +iv = z+b \bar z$, where $z$ traverses the unit circle exactly once.
After substituting $Re(z)= \cos \theta, Im(z) = \sin \theta$ and some algebra I get,
$ (1): \ v^2 + \frac{(1-b)^2 u^2}{(1+b)^2} = 1-b ^2$ for $ b \neq 1,-1$ and $(2): \ w=2Rez, (3): \ w= -2Imz$ for $ b=1$ and $b=-1$ respectively.
If it weren't for the fact that $b$ is complex I would say that $(1)$ is just a conic section. Can someone help me finish this off/ tell me where I've gone wrong? Many thanks!
 A: Writing $b = p + iq$ for $p, q \in \mathbb{R}$, we can expand the expression as
\begin{align*}
e^{i\theta} + be^{-i\theta}
&= (\cos\theta + i\sin\theta) + (p+iq)(\cos\theta-i\sin\theta) \\
&= ((1+p)\cos\theta + q\sin\theta) + i((1-p)\sin\theta + q\cos\theta).
\end{align*}
So if we write $x + iy = e^{i\theta} + be^{-i\theta}$ for real $x, y$, we are led to the system of linear equations
$$ \begin{pmatrix} x \\ y \end{pmatrix}
= \begin{pmatrix} 1+p & q \\ q & 1-p \end{pmatrix} \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix}. \tag{*}$$
If you are familiar with spectral theorem, this allows to deduce that the locus is a possibly degenerate ellipse whose major and minor axes are determined by the eigenvalue/eigenvector pairs of the 2 by 2 matrix appearing in $\text{(*)}$.
Otherwise, we can still tackle this problem by more elementary means. We have two cases:


*

*If $|b| \neq 1$ so that the 2 by 2 matrix appearing in $\text{(*)}$ is invertible, then solving $\text{(*)}$ to obtian
$$ \begin{pmatrix} \cos\theta \\ \sin\theta \end{pmatrix} = \frac{1}{1-|b|^2} \begin{pmatrix} 1-p & -q \\ -q & 1+p \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$
and then plugging this to the relation $\cos^2\theta + \sin^2\theta = 1$, we end up with
$$ ((1-p)x - qy)^2 + (-qx + (1+p)y)^2 = (1-|b|^2)^2. $$
This is the equation of the locus, and it is indeed an ellipse.

*If $|b| = 1$, then write $b = e^{i\alpha}$ and note that
$$ e^{i\theta} + be^{-i\theta}
= e^{i\alpha/2} \bigl( e^{i(\theta-\alpha/2)} + e^{-i(\theta-\alpha/2)} \bigr)
= 2 e^{i\alpha/2} \cos(\theta-\alpha/2). $$
So the locus is the line segment $[-2, 2]$ rotated around $0$ by $\alpha/2$ radian.
