Consider complex numbers $z_1 = 1 + i, z_2 = 1 – 3i, z_3 = 4 + i$ and complex variable $z$. Knowing that there exist complex numbers $z_4, z_5, z_6$ such that $\dfrac{z_4 – z_2}{z_4 – z_3}, \dfrac{z_5 – z_3}{z_5 – z_1}$ and $\dfrac{z_6 – z_1}{z_6 – z_2}$ are real numbers, while $\dfrac{z – z_4}{z_2 – z_3}, \dfrac{z – z_5}{z_3 – z_1}$ and $\dfrac{z – z_6}{z_1 – z_2}$ are imaginary numbers, determine the minimum value of $T = |z – z_4|^2 + |z – z_5|^2 + |z – z_6|^2$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
█▓▒░(°◡°)░▒▓█ I wish I was kidding sometimes.
First of all, it should be obvious that for all real numbers $x \not\in \{0; 1\}$, $\dfrac{1}{1 – x}$ and $\dfrac{1}{\dfrac{1}{x} – 1}$ are also real. In this case, $\dfrac{z_4 – z_3}{z_2 – z_3}, \dfrac{z_5 – z_1}{z_3 – z_1}, \dfrac{z_6 – z_2}{z_1 – z_2} \in \mathbb R$ and $\dfrac{z_4 – z_2}{z_2 – z_3}, \dfrac{z_5 – z_3}{z_3 – z_1}, \dfrac{z_6 – z_1}{z_1 – z_2} \in \mathbb R$.
Should I just say that $\dfrac{z_4 – \dfrac{z_2 + z_3}{2}}{z_2 – z_3}, \dfrac{z_5 – \dfrac{z_3 + z_1}{2}}{z_3 – z_1}, \dfrac{z_6 – \dfrac{z_1 + z_2}{2}}{z_1 – z_2}$ are real numbers?… I'll keep them here for now.
Furthermore, since $|z_2 – z_3| = 5, |z_3 – z_1| = 3$ and $|z_1 – z_3| = 4$, (if you listen carefully, you can hear the ancient Egyptians quaking in their tombs), $$T = |z – z_4|^2 + |z – z_5|^2 + |z – z_6|^2 = -\left[25 \times \left(\dfrac{z – z_4}{z_2 – z_3}\right)^2 + 9 \times \left(\dfrac{z – z_5}{z_3 – z_1}\right)^2 + 16 \times \left(\dfrac{z – z_6}{z_1 – z_2}\right)^2\right]$$
Other miscellaneous observations include $\Im(z_5) = 1, \Re(z_6) = 1$ and $$\dfrac{\Re(z_4) – \dfrac{5}{2}}{-3} = \dfrac{\Im(z_4) + 1}{-4} \implies z_4 = \left(3n + \dfrac{5}{2}\right) + (4n – 1)i \ (n \in \mathbb R)$$
Anyhow, that's all for now, I'll probably come back to this later.
As always, thanks for reading, (and even more so if you could help), have a great tomorrow, everyone~
By the way, the choices were $\dfrac{72}{5}, 3, \dfrac{72}{25}$ and $\dfrac{18}{25}$.
Best Answer
$\underline{\textbf{Preliminary Results}}$
PR-1
Given any $3$ distinct complex numbers, $a,b,c$ such that
$\displaystyle \frac{a-b}{a-c} = s \in \Bbb{R_{\neq 0}}$
where $(b-c) = re^{i\theta}$.
Then there exists $t \in \Bbb{R_{\neq 0}}$ such that
$(a-b) = tre^{i\theta}.$
Further, $(a)$ is somewhere on the infinite line that passes through points $(b)$ and $(c).$
Proof
$(a-b) = s(a-c) = s(b-c) + s(a-b) \implies (a-b)(1-s) = s(b-c).$
Since $a,b,c$ distinct, $s \neq 0$ and $s \neq 1.$
Let $\displaystyle ~t = \frac{s}{1-s}.$
Then $(a - b) = t(b-c) = tre^{i\theta}.$
If $t > 0$, then the argument of $(a-b)$ matches the argument of $(b-c)$, within a modulus of $(2\pi)$.
If $t < 0$, then $(a-b) = [r \times (-t)] e^{i(\theta + \pi)}.$
PR-2
Given any $4$ distinct complex numbers, $a,b,c,d$ such that
$\displaystyle \frac{a-b}{c-d} = is ~: ~s \in \Bbb{R_{\neq 0}}$
where $(c-d) = re^{i\theta}$.
Then, $(a-b) = sre^{i(\theta + \pi/2)}.$
Further, $(a)$ is somewhere on the infinite line that passes through $(b)$ and is perpendicular to the line segment connecting points $c$ and $d$.
Proof
$(a-b) = is(c-d) = e^{i(\pi/2)}sre^{i\theta}.$
If $s > 0$, then the argument of $(a-b)$ equals $(\theta + \pi/2)$, within a modulus of $(2\pi)$.
If $s < 0$, then $(a-b) = [r \times (-s)] e^{i(\theta + \pi/2 + \pi)}.$
Using PR-1, you have that
$(z_4)$ is on the infinite line connecting $(z_2)$ and $(z_3).$
$(z_5)$ is on the infinite line connecting $(z_3)$ and $(z_1).$
$(z_6)$ is on the infinite line connecting $(z_1)$ and $(z_2).$
Using PR-2, you have that
$(z)$ is on the infinite line that passes through $(z_4)$ and is perpendicular to the line connecting $(z_2)$ and $(z_3).$
$(z)$ is on the infinite line that passes through $(z_5)$ and is perpendicular to the line connecting $(z_3)$ and $(z_1).$
$(z)$ is on the infinite line that passes through $(z_6)$ and is perpendicular to the line connecting $(z_1)$ and $(z_2).$
Here, it helps to visualize the situation. $\triangle(z_1z_2z_3)$ is a right triangle with vertices at $(1 + i), (1 - 3i),$ and $(4 + i)$.
Clearly, $z_4, z_5, z_6$ are inside line segments $(\overline{z_2z_3}), (\overline{z_1z_3}),$ and $(\overline{z_1z_2})$, respectively.
$z$, is located somewhere inside the triangle, and the line segments connecting $z$ with each of $z_4, z_5, z_6$ form perpendiculars to the triangle's sides.
So, the problem can be solved by assuming that $z$ has Cartesian coordinates $(x,y)$, with $1 < x < 4, ~-3 < y < 1$. Based on the $(x,y)$ coordinate, the square of the distances between $z$ and $z_4, z_5,$ and $z_6$ will be established.
Then, $(x,y)$ will be chosen to minimize these distances.
The distance between $z$ and $z_6$ will be $(x - 1)$.
The distance between $z$ and $z_5$ will be $(1 - y)$.
The slope of $(\overline{z_2z_3})$ is $(4/3).$
Therefore, the slope of $(\overline{zz_4})$ must be $(-3/4).$
Let the Cartesian coordinates of $(z_4)$ be given by $(e,f).$
Then, $\displaystyle ~\frac{y-f}{x-e} = -\frac{3}{4}~~ \text{and}~~ \frac{1-f}{4-e} = \frac{4}{3}.$
This implies that
$16 - 4e = 3 - 3f \implies 4e - 3f = 13.$
$-3x + 3e = 4y - 4f \implies 3e + 4f = 3x + 4y.$
This resolves to
$$e = \frac{9x + 12y + 52}{25}, ~~ f = \frac{12x + 16y - 39}{25}. \tag1 $$
Calculating
$$(x - e)^2 = \frac{256x^2 + 144y^2 + 2704 - 384xy - 1664x + 1248y}{625}.$$
$$(y - f)^2 = \frac{144x^2 + 81y^2 + 1521 - 216xy - 936x + 702y}{625}.$$
Therefore,
$$(x-e)^2 + (y-f)^2 = \frac{16x^2 + 9y^2 + 169 - 24xy - 104x + 78y}{25}.$$
Then, squaring the distance from $(x,y)$ to each of $z_6$ and $z_5$ and adding them gives
$$\frac{(25x^2 - 50x + 25) + (25y^2 - 50y + 25)}{25}$$
$$= \frac{25x^2 + 25y^2 + 50 - 50x - 50y}{25}.$$
Combining this with $(x-e)^2 + (y-f)^2$ gives
$$\frac{41x^2 + 34y^2 + 219 - 24xy - 154x + 28y}{25}. \tag2 $$
So, with $x$ constrained by $1 < x < 4$
and $y$ constrained by $-3 < y < 1$,
the challenge is to choose $x,y$ so as to
minimize the numerator in (2) above.
Let $D(x,y)$ denote the numerator in (2) above.
Holding $x$ fixed, you have that
$\displaystyle \frac{\partial[D(x,y)]}{\partial y} = 68y - 24x + 28.$
Holding $y$ fixed, you have that
$\displaystyle \frac{\partial[D(x,y)]}{\partial x} = 82x - 24y - 154.$
In both of the equations above, it is immediate that the partial 2nd derivatives will be positive. So, computing $x,y$ to zero out both derivatives should produce the minimum.
This is achieved at $$(x,y) = \left(\frac{49}{25}, \frac{7}{25}\right).$$
Using (1) above, these values for $(x,y)$ result in
$$(e,f) = \left(\frac{73}{25}, \frac{-11}{25}\right).$$
So, the minimum sum of the squares of the distances is
$$\left(\frac{49}{25} - 1\right)^2 + \left(1 - \frac{7}{25}\right)^2 + \left(\frac{73-49}{25}\right)^2 + \left(\frac{7 + 11}{25}\right)^2$$
$$= \frac{72}{25}.$$
Addendum I should point out some things that will make a difference in a timed test.
Normally, in a multiple choice setting, I would simply assume that the preliminary results were true, just to save time. This means that I would proceed on that basis even though I wasn't sure that the conclusions were valid.
The choice of $(x,y)$ resulted in the square of the distance from $(x,y)$ to the hypotenuse to exactly equal the sum of the squares of the distances, to the two legs of the right triangle.
My knowledge of Real Analysis and Geometry is shallow in this area. The result didn't surprise me. However, I suspect that in the classroom setting, the student is supposed to know this in advance, and use this concept to shortcut the computations of $(x,y)$ and $(e,f)$.
In fact, the key idea would be to use this fact, compute $(x,y)$ alone (if possible), and then ignore $(e,f)$. You would then simply take the sums of the squares of $(x,y)$ to the two legs of the right triangle, and then double it.
There is an additional shortcut achievable by meta-cheating. It is clear from the initial visualization that the only plausible choices will be $(3)$ and $~\displaystyle \frac{72}{25}.$
So, you could bypass all other Math, and simply explore which of those two values results in the sum of the squares of $(x,y)$ to the two legs of the triangle equaling the square of the distance of $(x,y)$ to the hypotenuse.
In the meta-cheating world, this consideration immediately makes it game-over, although you do bypass the educational value of the problem.