Suppose that a, b, and c are complex numbers such that |a|=|b|=|c|=1 and a+b+c=0. Prove that |a-b|=|b-c|=|c-a| and discuss the geometric meaning.

The hint I received was to set a=1.....but I don't see where that is going.

Question

Suppose that a, b, and c are complex numbers such that |a|=|b|=|c|=1 and a+b+c=0. Prove that |a-b|=|b-c|=|c-a| and discuss the geometric meaning.

The hint I received was to set a=1.....but I don't see where that is going.

kirbykirby · Answer

I am almost done typing.. I'm sorry this is kind of long!!

kirbykirby · Answer

This is a bit long.. Maybe someone else has a shorter solution
Let: $a=x_1+iy_1$
$b=x_2+iy_2$
$c=x_3+iy_3$

Then:
$|a|=\sqrt{x_1^2+y_1^2}=1\implies x_1^2+y_1^2=1$
$|b|=\sqrt{x_2^2+y_2^2}=1\implies x_2^2+y_2^2=1$
$|c|=\sqrt{x_3^2+y_3^2}=1 \implies x_3^2+y_3^2=1 $

and:

$a+b+c=0 \
\implies (x_i+iy_1)+(x_2+iy_2)+(x_3+iy_3)=0\
\implies (x_1+x_2+x_3)+i(y_1+y_2+y_3)=0 +i0\
\~ \ \implies x_1+x_2+x_3=0 ~\text{    and    }~y_1+y_2+y_3=0 $

Which implies:
(I'll call these equations by$ (*)$
$x_1=-(x_2+x_3)\
x_2=-(x_1+x_3)\
x_3=-(x_1+x_2)$
and the same idea for the $y$'s

So:
$$ \begin{align}|a-b|&=|(x_1+iy_1)-(x_2+iy_2)|\&=|(x_1-x_2)+i(y_1-y_2)| \&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\&=\sqrt{x_1^2-2x_1x_2 +x_2^2+y_1^2-2y_1y_2+y_2^2}\&=\sqrt{\underbrace{x_1^2+y_1^2}_{=1} +\underbrace{x_2^2+y_2^2}_{=1}-2(x_1x_2+y_1y_2 ) }\&=\sqrt{2-2(x_1x_2+y_1y_2)}\end{align}$$

Analogously, you will obtain the following: 
$$|b-c|=\sqrt{2-2(x_2x_3+y_2y_3)} $$$$ |c-a|=\sqrt{2-2(x_3x_1-y_3y_1})$$

Now if you substitute every single x and y with the formulas in $(*)$:

Just focusing on the term: $x_1x_2$ $$ \begin{align}x_1x_2&= [-(x_2+x_3)][-(x_1+x_3)]\&=(x_2+x_3)(x_1+x_3) \&= x_2x_1+x_2x_3+x_3x_1+x_3^2 \end{align}$$
Thus, $y_1y_2= y_2y_1+y_2y_3+y_3y_1+y_3^2$

In a similar way, you will find:
$x_2x_3=x_1^2+x_1x_2+x_3x_1+x_3x_2$
$x_3x_1=x_1x_2+x_1x_3+x_2^2+x_2x_3$ 

Thus:
$$ |a-b|=\sqrt{2-2[(x_2x_1+x_2x_3+x_3x_1+x_3^2) +(y_2y_1+y_2y_3+y_3y_1+y_3^2)]}$$  
$$ |b-c|=\sqrt{2-2[(x_1^2+x_1x_2+x_3x_1+x_3x_2) +(y_1^2+y_1y_2+y_3y_1+y_3y_2)]}$$  
$$ |c-a|=\sqrt{2-2[(x_1x_2+x_1x_3+x_2^2+x_2x_3) +(y_1y_2+y_1y_3+y_2^2+y_2y_3)]} $$

kirbykirby · Answer

if you combine $x_1^2+y_1^2=1$
$x_2^2+y_2^2=1$
$x_3^2+y_3^2=1$

Then all the expression are the same

kirbykirby · Answer

Sorry I made a typo here: $$|c-a|=\sqrt{2-2(x_3x_1+y_3y_1})$$

zarkon · Answer

it is much simpler to use the hint provided (let a=1) 

rotations of the circle (the vectors) does not change their length nor the length of their differences

kirbykirby · Answer

It's still valid though what I wrote, I hope?

zarkon · Answer

probably ...i didn't read through it ;)

I was working on other stuff

myininaya · Answer

Wouldn't get a counterexample if we let (a,b,c)=(1,1,-1)?

myininaya · Answer

oops the sum is 0 not 1

myininaya · Answer

my bad

zarkon · Answer

I get the lengths of the differences to be $\sqrt{3}$

zarkon · Answer

you can do the entire problem by using geometry on the unit circle

anonymous · Answer

thank yall!