How exactly do I "show" that $ z = ae^{i\alpha} + be^{-i\alpha} $ represents an ellipse?

Question

Here is an equation and I want to know how I can show that this represents an ellipse ($a$ and $b$ are complex constants and $\alpha$ is a real variable): $$ z = ae^{i\alpha} + be^{-i\alpha} $$

I wanna learn how a mathematician would approach a problem like this, like where would you start for example. I have no idea where to even begin with problems like these.

Answer 1

To better understand complex mappings, it is often usefull to either split an equation into real and imaginary part, or to transform it entirely into polar coordiantes. Let's try the second idea:

Write $a = R_1 e^{i\phi_1}$, $b=R_2e^{-i\phi_2}$, then you get:

$$z(\alpha) = ae^{i\alpha}+be^{-i\alpha} = R_1e^{i(\phi_1+\alpha)}+R_2e^{-i(\phi_2+\alpha)}$$

To further simplify this, we want both terms to have the same argument. Notice that the image $\{z(\alpha) : \alpha\in\mathbb{R}\}$ traced out by $z(\alpha)$ for $\alpha\in\mathbb{R}$ does not change if we reparametrize the curve, i.e. we can introduce a new parametrization $\beta$ via $\alpha = \beta - \underbrace{\frac{\phi_1+\phi_2}{2}}_{=:\phi}$, which transforms above formula to the following:

$$z(\beta ) = R_1 e^{i(\beta+\frac{\phi_1}{2}-\frac{\phi_2}{2})} + R_2e^{-i(\beta -\frac{\phi_1}{2}+\frac{\phi_2}{2})} = R_1e^{i(\beta+\phi)}+R_2e^{-i(\beta-\phi)} = e^{i\phi}(R_1e^{i\beta}+R_2e^{-i\beta}),$$ where $\phi = \frac{\phi_1+\phi_2}{2}$ is just some constant, meaning $e^{i\phi}$ is a rotation by the angle $\phi$.

Now you can use Euler's identity $e^{i\beta} = \cos\beta + i\sin\beta$ to further simplify this, using that $\cos\beta = \cos(-\beta)$ and $-\sin\beta =\sin(-\beta)$. In the end, this should yield

$$z(\beta) = e^{i\phi}((R_1+R_2)\cos\beta + i(R_1-R_2)\sin\beta),$$

Basically (for $R_1 \neq\pm R_2$) the term $(R_1+R_2)\cos\beta + i(R_1-R_2)\sin\beta)$ just describes an standard ellipse centerd at $0$ with radii $R_1+R_2$ and $R_1-R_2$ (since it satisfies an equation of the form $\frac{x^2}{\lambda}+\frac{y^2}{\delta}=1$ for $\lambda,\delta>0$), while the factor $e^{i\phi}$ in front of it rotates this ellipse around the origin by $\phi$.

If you prefer cartesian coordinates: $$z(\beta) = \left(\begin{array}{cc}\cos\phi & -\sin\phi \\ \sin\phi & \cos\phi\end{array}\right)\left(\begin{array}{c}(R_1+R_2)\cos\beta \\ (R_1-R_2)\sin\beta\end{array}\right)$$

Answer 2

I wanna learn how a mathematician would approach a problem like this

I will try to answer this part of the question. There are various ways a geometric object is defined. Showing that something represents a given geometric object $G$ means matching any of its definitions

Take the example of a circle of center $C$ and radius $R$. It is usually defined as the set of all points in the plane whose distance to $C$ is equal to $R$. Then, using a coordinates system, we can obtain an equation of the circle: if $C(a,b)$, then a point $M(x,y)$ is on the circle if and only if

$$(x-a)^2+(y-b)^2=R^2$$

Let's call this the standard equation of the circle. Note that we could also have used it as a definition since it is equivalent to the original definition. To show that something represents a circle, you can either show it is the set of points equidistant to a fixed point, or show that you can rewrite the data you have into the standard equation of a circle. The exact strategy depends on the problem.

Now, there is not a single method to show that something represents a geometric object. Here are a few examples:

1. to show that the equation $x^2-2x+y^2+2y-2=0$ represents a circle, you can complete the square:

$$x^2-2x+y^2+2y-2=0 \iff (x-1)^2-1+(y+1)^2-1-2=0 \iff (x-1)^2+(y+1)^2=2^2$$

so the equation represents a circle of center $(1,-1)$ and radius $2$ (by comparison with the standard form).

2. The set of all points $(x,y)$ such that $x=2\cos(t)$ and $y=1+2\sin(t)$, $t\in\mathbb R$, is a circle, since $x^2+(y-1)^2=4\cos^2(t)+4\sin^2(t)=4$.

3. You can use a rotation of coordinate axes to show that $3x^2-2\sqrt{3}xy+y^2+8x+8\sqrt{3}y=0$ is a hyperbola by changing the equation into the standard equation of a hyperbola

4. If $a=2+3i$, the equation $\overline{z}+\overline{a}z=1$ represents a line. Indeed, write $z=x+iy$ with $x,y$ real numbers. Then the equation is equivalent to

$$ (2+3i)(x-iy)+(2-3i)(x+iy)=1 \iff 4x+6y=1 $$

which is a standard equation for a line.

Answer 3

$z=ae^{it}+be^{-it}, a,b,t\in \Re \implies x+iy=(a+b) \cos t+i(a-b)\sin t$

$\implies x=(a+b) \cos t, y=(a-b)\sin t \implies \cos t=\frac{x}{a+b}, \sin t=\frac{y}{a-b}$

Squaring and adding las two we get $\frac{x^2}{(a+b)^2}+\frac{y^2}{(a-b)^2}=1,$ which is and ellipse if $a\ne b$.

If $b=a$ then $x+iy=2a \cos t,y=0$, it represents a line sefment on $x$axis from $x=-2a,$ to $x=2a$.