# Thread: Complex plane area problem

1. ## Complex plane area problem

A friend gave me this question as part of a puzzle but I am unfamiliar with how to approach. Help? Thanks

In the complex plane, z, z^2, z^3 form, in some order, three vertices of a non-degenerate square. Determine, in lowest terms, the harmonic mean, of the 3 possible areas of these squares.

2. ## Re: Complex plane area problem

I might be missing something but it seems like every square, other than one of zero area, that meets these requirements will have vertices at

$i, -1, -i, 1$

$\large z = e^{i \pi \frac{2k+1}{2}},~k\in \mathbb{Z}$

all these squares have sides of length $s=\sqrt{2}$ and thus area of $A=2$

The harmonic mean is thus of course 2

3. ## Re: Complex plane area problem

it could also have vertices at $\displaystyle -1+i, -2i,3-i,2+2i$

4. ## Re: Complex plane area problem

how did you come up with that?

5. ## Re: Complex plane area problem

The three cases mentioned correspond to whether the diagonal of the square is the segment joining

case 1. $\displaystyle z$ to $\displaystyle z^2$

case 2. $\displaystyle z$ to $\displaystyle z^3$

or

case 3. $\displaystyle z^2$ to $\displaystyle z^3$

in case 1, the line $\displaystyle (z,z^3)$ must be perpendicular to the line $\displaystyle (z^2,z^3)$

$\displaystyle z^3-z^2=i(z^3-z)$

similarly for the other cases

we get three possible values for the area: $\displaystyle 5/8, 2, 10$

7. ## Re: Complex plane area problem

Thanks for your help. I get a harmonic mean of 15/11.

8. ## Re: Complex plane area problem

I don't think 15/11 is the right answer. I get $\displaystyle \frac{50}{39}$

There are 3 possible areas but there actually are 5 squares,
two with area $\displaystyle 10$
one with area $\displaystyle 2$
two with area $\displaystyle 5/8$

We should be taking the harmonic mean of the list $\displaystyle \{10,10,2,5/8,5/8\}$

Use this fact from complex numbers:

$\displaystyle u,v$ are orthogonal and have the same magnitude if and only if $\displaystyle u=\pm i v$

9. ## Re: Complex plane area problem

Based on the way the original question is stated, I think its just the three areas, not the areas of the five squares.

10. ## Re: Complex plane area problem

Sorry, you are right, it is 15/11