Too Complex to Solve!!

Here Answer is : $\displaystyle{3({ \left| { Z }_{ 1 } \right| }^{ 2 }+{ \left| { Z }_{ 2 } \right| }^{ 2 }+{ \left| { Z }_{ 3 } \right| }^{ 2 })}$ .

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The expression equals $3(|z_1|^2+|z_2|^2+|z_3|^2) - |z_1+z_2+z_3|^2$ . So the best we can do is to choose the $z_i$ so that they sum to 0, whence the answer is $3(4+9+16)=\fbox{87}$ .

(We can so choose the $z_i$ because there is a triangle with sides $2,3,4$ . This is an important point--if we changed the constant values of the $|z_i|$ , the problem's answer might not be 3 times the sum of those squares anymore.)

Let $z_1 = x_1 +iy_1 , z_2 = x_2 +iy_2$ & $z_3 = x_3 +iy_3$ .

Then $x_{1}^{2} +y_{1}^{2} = 4, x_{2}^{2} +y_{2}^{2} = 9$ & $x_{3}^{2} +y_{3}^{2} = 16$

And, $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 =$ $2(x_{1}^{2} +y_{1}^{2} +x_{2}^{2} +y_{2}^{2}+ x_{3}^{2} +y_{3}^{2}) -$ $2(x_{1} x_{2} + y_{1} y_{2} + x_{2} x_{3} + y_{2} y_{3} + x_{3} x_{1} + y_{3} y_{1})$ .

$\implies$ $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 = 58 -$ $2(x_{1} x_{2} + y_{1} y_{2} + x_{2} x_{3} + y_{2} y_{3} + x_{3} x_{1} + y_{3} y_{1})$ .

So in order to find the maximum value of $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2$ , we have to find minimum value of $x_{1} x_{2} + y_{1} y_{2} + x_{2} x_{3} + y_{2} y_{3} + x_{3} x_{1} + y_{3} y_{1}$ .

Since we know that square of any real number is greater than or equal to 0. So we can use $(x_1 +x_2 +x_3)^2 +(y_1 +y_2 +y_3)^2 \ge 0$

$\implies$ $x_{1}^{2} +y_{1}^{2} +x_{2}^{2} +y_{2}^{2}+ x_{3}^{2} +y_{3}^{2} +$ $2(x_{1} x_{2} + y_{1} y_{2} + x_{2} x_{3} + y_{2} y_{3} + x_{3} x_{1} + y_{3} y_{1}) \ge 0$ .

$\implies$ $(x_{1} x_{2} + y_{1} y_{2} + x_{2} x_{3} + y_{2} y_{3} + x_{3} x_{1} + y_{3} y_{1}) \ge -\frac{29}{2}$

Hence $|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \le 58 - 2\times (-\frac{29}{2})$

or $\boxed{|z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \le 87}$

I just don't like complex numbers it's my weakest topic. Hence I am here presenting an algebric approach to this question with a little bit of calculus.

Let ${z}_{1} = 2(cos{ \theta }_{ 1 }+isin{ \theta }_{ 1 })$

${z}_{2} = 3(cos{ \theta }_{ 2 }+isin{ \theta }_{ 2 })$

${z}_{3} = 4(cos{ \theta }_{ 3 }+isin{ \theta }_{ 3 })$

So the expression that needs to be maximised get's converted into :

$S = 58-(12cos({ \theta }_{ 1 }-{ \theta }_{ 2 })+24cos({ \theta }_{ 2 }-{ \theta }_{ 3 })+16cos({ \theta }_{ 3 }-{ \theta }_{ 1 }))$

Now take ${\theta}_{1}-{\theta}_{2}=x,{\theta}_{2}-{\theta}_{3}=y$

$S = 58-(12cos(x)+24cos(y)+16cos(x+y))$

After some simplification we have :

$S = 58 -( (12+16cos(y))cos(x)+(16siny)sin(x) + 24cos(y) )$

Remember that we have to maximise $S$

and hence it comes down to minimizing

$f(y) = (12+16cos(y))cos(x)+(16siny)sin(x) + 24cos(y)$

and $x$ and $y$ are independent variables and minimum value of $acos(x)+bsin(x)$ is $-\sqrt{{a}^{2}+{b}^{2}}$

Hence we are minimizing :

$f(y) = 24cos(y)-\sqrt { 400+384cos(y) }$

differentiate and equate it to zero to get :

$cos(y) = -\dfrac{21}{24}$

Finally ${f(y)}_{min} = 29$

$\boxed{{S}_{max} = 87}$

Finally accomplished, using partial derivatives,

I present a partial derivative approach, which is awesome in certain situations like finding tangent or normal for implicit equations instead of using normal derivative or to easily find out rate of change along any 3-D direction or ofcourse as here to extremise functions that depend on more than one varriable

As the situation is invariant under rotation, let us fix one of the points as (2,0) (the inner most one ofcourse, my wish)

So we have

$z_1$ = 2+i

$z_2$ = 3(cos(y) + i sin(y))

$z_3$ = 4(cos(z) + i sin(z))

so inputting in the expression using distance formula

we have

58-(12cos(y) + 24cos(z) + 16 cos( y-z)) = S which has to be maximised

so we have to minimise

12cos(y) + 24cos(z) + 16 cos( y-z)=f

so let us partial differentiate with respect to x and y set each = 0

to get

$\frac { \delta f }{ \delta y } =0=-12sin(y)-16sin(y-z)----(a)\\ \\ \frac { \delta f }{ \delta z } =0=-24sin(z)+16sin(y-z)----(b)\\ \\ combining\quad both\quad we\quad also\quad get\\ \\ \frac { -1 }{ 2 } sin(y)\quad =\quad sin(z)----(c)\\ \\ now\quad from\quad (a)\quad we\quad get,\\ \\ \frac { -3 }{ 4 } sin(y)\quad =\quad sin(y-z)\quad =sin(y)cos(z)\quad -\quad cos(y)sin(z)\quad \\ \\ substitung\quad (b)\quad ,\quad and\quad simplifying(squaring\quad after\quad transposing)\quad we\quad get,\\ \\ cos(z)\quad =\quad \frac { -7 }{ 8 } \quad from\quad which\quad we\quad also\quad get\quad cos(y)\quad =\quad \frac { 1 }{ 4 } \quad cos(z-y)\quad =\quad \frac { -11 }{ 16 } \\ \\ substituting,\quad we\quad have\quad -29\quad as\quad the\quad minimum\quad value\\$

so we have 58+29 = 87 which is the answer

its a shame i took so long,