A faster way to find the complex locus of $ | z^* + 2i | + | z | = a $

Question

I need to find the equation of the locus of $ z = x + iy $ which satisfies

$$ | z^* + 2i | + | z | = a $$

for some real constant $ a $, where $ z^* $ is the conjugate, $ x - iy $.

I already solved this problem by simply substituting these forms in and going through the extremely lengthy algebra (which involves expanding to an equation with 14 terms up to quartic degree) and there must be a quicker way.

Intuitively I can see that the locus equation is saying that the sum of two distances from $ 0 $ and $ 2i $ is constant, suggesting an ellipse with foci at these points. Is there a way to get to the equation from here? What about only in the special cases $ a = 4 $ and $ a = 2 $?

For reference, the equation is - verified with this plotter

$$ a^2 x^2 + (a^2 - 4)(y - 1)^2 = \frac{1}{4} a^2 (a^2 - 4). $$

Thanks for any help!

Answer 1

From wikipedia:

An ellipse can be defined geometrically as a set or [[locus of points]] in the Euclidean plane: : Given two fixed points$F_1, F_2$ called the foci and a distance $2a$ which is greater than the distance between the foci, the ellipse is the set of points $P$ such that the sum of the distances $|PF_1|,\ |PF_2|$ is equal to $2a:E = \{P\in \mathbb{R}^2 \,\mid\, |PF_2| +|PF_1 | = 2a \}$

$$|z^*+2i|+|z|=a$$

$$|z-2i|+|z|=2 \left( \frac{a}{2} \right)$$

That is the foci are $(0,2)$ and $(0,0)$. The major axis is the $y$-axis.

The center of the ellipse is $(0,1)$.

$$b^2 = \left( \frac{a}{2}\right)^2-\left( \frac{2-0}{2}\right)^2=\frac{a^2-4}4 $$

Hence the equation is

$$\frac{4x^2}{a^2-4}+\frac{4(y-1)^2}{a^2}=1$$

$$4a^2x^2 + 4(a^2-4)(y-1)^2 = a^2(a^2-4)$$

Answer 2

Assuming the question is how to bring the complex equation to the canonical cartesian form:

$$ \require{cancel} \begin{align} |\bar z+2i|^2 &= \left(a - |z|\right)^2 \\ \left(\bar z+2i\right)\left(z - 2i\right) &= a^2 - 2a |z| + |z|^2 &&\;\;\;\;\text{using: }|\bar z+2i|^2=(\bar z+2i)\overline{(\bar z+2i)} \\ \cancel{|z|^2} + 4 - 4 \Im(z) &= a^2 - 2a |z| + \cancel{|z|^2} &&\;\;\;\;\text{using: } z-\bar z = 2i\, \Im(z) \\ 2a\sqrt{x^2+y^2} &= 4 y + a^2 - 4 &&\;\;\;\;\text{using: } |z| = \sqrt{x^2+y^2}, \;\Im(z) = y \end{align} $$

Squaring the last equality and regrouping the terms in $x,y$ gives the equation as posted.