What is the sum of complex solutions?

[unscientific]

Homework Statement

Let z₁, ... z_n be the set of n distinct solutions to the equation

zⁿ = a

where a is a complex number.

(a) By considering distinct solutions as the sides of a polygon in an Argand diagram show that these sum to zero.(b) Hene find the sum of the squares of these solutions. For the case n = 5 sketch the polygon traced out by these successive squared values in the Argand plane.

The Attempt at a Solution

Of course this can be easily solved by doing summation of geometric series, but this isn't a typical question...

I managed to show that the sum of sides = 0 algebraically. This can be shown by using vectors as well (red arrows) that starting from point z₁ you will arrive back at z₁, implying the overall change = 0.

But, how do i relate each side of the polygon to one solution? Is there a bijection somewhere?

 

[HallsofIvy]

For any $a_1$, $a_2$, .. $a_n$, $(z- a_1)(z- a_2)...(z- a_n)= z^n- ... - (a_1+ a_2+...+ a_n)z+ a_1a_2...a_n$. In particular, if $z^n= a$, so that the coefficient of z is 0, it follows that the sum of roots is 0. That doesn't require complex numbers.

 

[unscientific]

For any $a_1$, $a_2$, .. $a_n$, $(z- a_1)(z- a_2)...(z- a_n)= z^n- ... - (a_1+ a_2+...+ a_n)z+ a_1a_2...a_n$. In particular, if $z^n= a$, so that the coefficient of z is 0, it follows that the sum of roots is 0. That doesn't require complex numbers.

Do you mean coefficient of z^n-1?

Hmm that's a very nice way to prove it.

But how does each term in the bracket correspond to a side?

and part (b) what does the square of that have to do with anything at all?

 

[haruspex]

The idea is that if the vertices of the polygon are at x_k+iy_k then the roots are the differences between consecutive vertices. But the question seems a bit backwards to me. How do you show that you can construct a closed polygon that way? Seems to me you first need to prove that roots sum to zero, e.g. in the way Halls showed.
Squaring a complex number doubles its angle to the positive real axis. Without worrying about what it does to the lengths of the sides (other than that they will still be all the same length), what does that do to the polygon?

 

[unscientific]

The idea is that if the vertices of the polygon are at x_k+iy_k then the roots are the differences between consecutive vertices. But the question seems a bit backwards to me. How do you show that you can construct a closed polygon that way? Seems to me you first need to prove that roots sum to zero, e.g. in the way Halls showed.
Squaring a complex number doubles its angle to the positive real axis. Without worrying about what it does to the lengths of the sides (other than that they will still be all the same length), what does that do to the polygon?

The vertices of the polygon are the roots. Now the problem is to relate the vertices to the sides... which is the vector difference between any two roots..

Squaring a complex number merely doubles the angle and forces all vertices to the Re(x)>0 side..

 

[haruspex]

The vertices of the polygon are the roots.

No, it says "By considering distinct solutions as the sides of a polygon"

Squaring a complex number merely doubles the angle

Right, so it doubles the angle between consecutive edges. How would you describe the effect on the polygon as a whole.

and forces all vertices to the Re(x)>0 side..

Why?

 

[unscientific]

No, it says "By considering distinct solutions as the sides of a polygon"

Right, so it doubles the angle between consecutive edges. How would you describe the effect on the polygon as a whole.

Why?

a) how can you simply "consider sides as distinct solutions" when they are not? i was trying to show the relation, starting from vertices as the roots..


b)the number of sides is reduced by half.. since now all positive solutions are "positive" and "negative" roots are indistinguishable after squaring them..

 

[haruspex]

a) how can you simply "consider sides as distinct solutions" when they are not?

I agree that 'considering' the sides as distinct solutions is rather glib. You have to show by some other means that they add up to zero, and only then can you be sure the sides will form a closed polygon. But that is what the question says.

b)the number of sides is reduced by half..

Only if n is even.

since now all positive solutions are "positive" and "negative" roots are indistinguishable after squaring them..

No, you can't use that sort of reasoning. These are complex numbers. Squaring doesn't make them positive in any sense.
Squaring the sides makes the polygon go around the origin twice.