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.
Best Answer
$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).