Graphing and describing the set $|z+2-3i|+|z-2+3i|\leq 10$.

Question

I need help finishing the following problem.

Problem: Describe the following set $S=\big\{z_{0}\in\mathbb{C}:|z_{0}+2-3i|+|z_{0}-2+3i|\leq 10\big\}$.

Note that we are to provide a graph too, and then answer the questions as to whether the given set is a domain and whether, or not, the set is bounded in $\mathbb{C}$. If I can get help describing the set, I can do the rest of the problem easily. Furthermore, this problem is a review question for an exam that I have coming up on Monday (2/13/2017), so any help is GREATLY appreciated.

In terms of simplifying everything (unless there is an easier way), we let $z=x+iy\in S$, for some $x,y\in\mathbb{R}$. First we plug $z=x+iy$ into the expression $|z+2-3i|+|z-2+3i|$ and then use the definition of the absolute value (or modulus) of a complex number. In terms of my work (note I'm going to omit some minor details since my query falls after the algebraic simplification below), we get:

$|z+2-3i|+|z-2+3i|=\sqrt{(x+2)^{2}+(y-3)^{2}}+\sqrt{(x-2)^{2}+(y+3)^{2}}\leq 10$.

Moving the right-hand square root to the right-hand side of the inequality above, by subtraction, we get:

$\sqrt{(x+2)^{2}+(y-3)^{2}}\leq 10-\sqrt{(x-2)^{2}+(y+3)^{2}}$.

Squaring both sides, collecting like terms and further simplifying, we now have:

$(x+2)^{2}+(y-3)^{2}\leq 100-20\sqrt{(x-2)^{2}+(y+3)^{2}}+(x-2)^{2}+(y+3)^{2}$

$8x-12y\leq 100-20\sqrt{(x-2)^{2}+(y+3)^{2}}$

$2x-3y\leq 25-5\sqrt{(x-2)^{2}+(y+3)^{2}}$.

Squaring both sides a second time to remove the square root, we get:

$4x^{2}+9y^{2}-12xy-100x+150y+625\leq 25\big((x-2)^{2}+(y+3)^{2}\big)$.

After simplifying the inequality directly above, we end up with:

$21x^{2}+12xy+16y^{2}\leq 300$.

[[ NOTE: An algebraic error existed in the last inequality directly above - this was edited with the aid of the solutions/comments below -- my confusion originated from the $12xy$ term. ]]

I can't figure out what to do now, as I've been looking over the equation of an ellipse in the complex plane, given by $E=\big\{z_{0}\in\mathbb{C}:|z_{0}-a|+|z_{0}-b|=c\big\}$, where $a,b,c$ are arbitrarily fixed constants in $\mathbb{C}$; in this case, $a=-2+3i=-b$ and $c=10~$ (as far as within the inequality that all points within $S$ satisfy). I worked the steps above twice already, and I came up with the same thing, so I think all my algebra is correct. However, where do I go from here? I feel I'm close, but I can't seem to describe the set given the last inequality derived above. As it is mentioned, any help is GREATLY appreciated in terms of whether I made an algebraic mistake, where to go from here, etc.

Answer 1

First off, let me mention that the inequality that you arrived at should be the reverse. It probably was a gross error. (EDIT: The error in the OP's work has been fixed.)

An ellipse is defined as the equation of the locus of a moving point such that the sum of its distances from two fixed points (the foci) remains constant and equal to the length of the major axis.

In case of the set you provided, the foci of the bounding ellipse are $2-3i$ and $-2+3i$ which are diametrically opposite points. Hence the major axis of the ellipse is along the line passing through the foci, i.e. $2y + 3x =0$. Accordingly, the minor axis will be perpendicular to this line and also pass through the origin, i.e. $3y - 2x = 0$. The fact that $10\ge |(2-3i)-(-2+3i)| = 2\sqrt{13}$ lends support to the observation that the bounding curve is an ellipse. Hence, as the other answers and comments have mentioned, we conclude that the set represents all points in the interior of the ellipse.

Let's now concentrate on the bounding ellipse, i.e. $|z-2+3i|+|z+2-3i| = 10$. If the major axis be $2a$, then $2a = 10 \implies a=5$. Distance between the foci is $2ae$, where $e$ is the eccentricity. Thus, $e = \frac{\sqrt{13}}{5}$. If the minor axis be $2b$, then $b = a\sqrt{1-e^2} \implies b = \sqrt{12}$. With all this information, we can write the equation of the ellipse as:

$$ \frac{X^2}{a^2}+\frac{Y^2}{b^2} = 1 $$

in a coordinate system with the axes being the lines $2y + 3x = 0$ and $3y-2x=0$ respectively. This the equation of the ellipse in the original system is:

$$ \frac1{b^2} \left(\frac{2y+3x}{\sqrt{13}}\right)^2 + \frac1{a^2} \left(\frac{3y-2x}{\sqrt{13}}\right)^2 = 1 $$

which should give you the equation of the bounding ellipse as $21x^2 + 12xy+ 16y^2 = 300$. This completes the analysis of the set.

Answer 2

$S=\big\{z_{0}\in\mathbb{C}:|z_{0}+2-3i|+|z_{0}-2+3i|\leq 10\big\} $

Immediately, this is an ellipse with foci at $(-2, 3)$ and $(2, -3)$ and the sum of the distances to the foci is $10$.

Also, in your calculations, when you have $2x-3y\leq 25-5\sqrt{(x-2)^{2}+(y+3)^{2}}$, I would continue $5\sqrt{(x-2)^{2}+(y+3)^{2}} \le 25-2x+3y $.

Squaring,

$\begin{array}\\ 25((x-2)^{2}+(y+3)^{2}) \le 4 x^2 - 12 x y - 100 x + 9 y^2 + 150 y + 625\\ \iff\\ 25(x^2-4x+4+y^2+6y+9) \le 4 x^2 - 12 x y - 100 x + 9 y^2 + 150 y + 625\\ \iff\\ 25x^2-100x+100+25y^2+150y+225 \le 4 x^2 - 12 x y - 100 x + 9 y^2 + 150 y + 625\\ \iff\\ 21x^2+ 12 x y+16y^2 \le 300\\ \end{array} $

(if my algebra is correct).

Answer 3

I think it's better to say the axis of this ellipse passes through a line made by $2-3i$ and $-2+3i$ with slope $\theta_0=\arctan\dfrac{-3}{2}$, so with a rotation $z\to ze^{i\theta_0}$ the rotated ellipse will be $$|ze^{i\theta_0}+2-3i|+|ze^{i\theta_0}-2+3i|=10$$ or $$|z+\sqrt{13}|+|z-\sqrt{13}|=10$$