Show that $z= Ae^{it}+Be^{-it}$ is an ellipse

Question

If $A$ and $B$ are complex parameters. Apparently the semi-axes should be |A| and |B|, and the orientation should be $\frac{1}{2}$arg(A)+arg(B). I rewrote the original expression as |A|$e^{at}$$e^{it}$+|B|$e^{bt}e^{-it}$ where a and b are the arguments of the parameters A and B. I then used Euler's formula and tried to rearrange using every trig formula I could find, to get the equation $z=e^{i\frac{a+b}{2}}(|A|cos(t)+i|B|sin(t))$ with no success...any help would be much appreciated!

Answer 1

Let $\alpha,\beta$ be the arguments of $A,B$, and let $\gamma$ be their half-sum and $\delta$ their half-difference. The equation reads

$$|A|e^{i(t+\alpha)}+|B|e^{-i(t+\beta)}$$

or

$$e^{i\gamma}(|A|e^{i(t-\delta)}+|B|e^{-i(t-\delta)}).$$

We can shift the parameter $t$ by $\delta$, and rotate the curve by $e^{-i\gamma}$. Now the equation reads

$$|A|e^{it}+|B|e^{-it}$$ or $$\begin{cases}x=|A|\cos t+|B|\cos t,\\y=|A|\sin t-|B|\sin t\end{cases}$$

which is an ellipse of half-axis $(|A|+|B|)$ and $(||A|-|B||)$