Unproved patterns

Question

Let $z_1,z_2,z_3\in\mathbb{C}$ such that $|z_1|=|z_2|=|z_3|=1, z_1+z_2+z_3=1$. Calculate $z_1^{2003}+z_2^{2003}+z_3^{2003}.$

Let's say $S_k=z_1^k+z_2^k+z_3^k$. I tried calculating $S_2,S_3,S_4,S_6$ so maybe I see a pattern.

$|z|=1\Leftrightarrow \bar{z}=\frac{1}{z}.~z_1+z_2+z_3=1\Rightarrow \bar{z_1}+\bar{z_2}+\bar{z_3}=1\Rightarrow \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=1\Rightarrow \mathbf{z_1z_2+z_1z_3+z_2z_3=z_1z_2z_3}.$

Also $z_1+z_2+z_3=1\Rightarrow (z_1+z_2+z_3)^2=1\Rightarrow \underbrace{z_1^2+z_2^2+z_3^2}_{S_2}+2\underbrace{(z_1z_2+z_1z_3+z_2z_3)}_{z_1z_2z_3}=1$ $\Rightarrow S_2+2z_1z_2z_3=1$

$\mathbf{\Rightarrow S_2=1-2z_1z_2z_3}$

$S_3=z_1^3+z_2^3+z_3^3=\underbrace{(z_1+z_2+z_3)}_1\big(\underbrace{z_1^2+z_2^2+z_3^2}_{S_2}-\underbrace{(z_1z_2+z_1z_3+z_2z_3)}_{z_1z_2z_3}\big)+3z_1z_2z_3$

$S_3=S_2-z_1z_2z_3+3z_1z_2z_3=S_2+2z_1z_2z_3\Rightarrow \mathbf{S_3=1}$

I will calculate an identity we will need for $S_4$:

$z_1z_2+z_1z_3+z_2z_3=z_1z_2z_3\Rightarrow (z_1z_2+z_1z_3+z_2z_3)^2=(z_1z_2z_3)^2$

$\Rightarrow (z_1z_2)^2+(z_1z_3)^2+(z_2z_3)^2+2z_1z_2z_3\underbrace{(z_1+z_2+z_3)}_1=(z_1z_2z_3)^2$

$\Rightarrow\mathbf{(z_1z_2)^2+(z_1z_3)^2+(z_2z_3)^2=(z_1z_2z_3)^2-2z_1z_2z_3}$

$S_4=z_1^4+z_2^4+z_3^4={\underbrace{(z_1^2+z_2^2+z_3^2)}_{S_2}}^2-2\underbrace{((z_1z_2)^2+(z_1z_3)^2+(z_1z_3)^2)}_{(z_1z_2z_3)^2-2z_1z_2z_3}$

$S_4=(1-2z_1z_2z_3)^2-2\big((z_1z_2z_3)^2-2z_1z_2z_3\big)=4(z_1z_2z_3)^2-4z_1z_2z_3+1-2(z_1z_2z_3)^2+4z_1z_2z_3$ $\Rightarrow \mathbf{S_4=1+2(z_1z_2z_3)^2}$

$S_6=z_1^6+z_2^6+z_3^6={\underbrace{(z_1^3+z_2^3+z_3^3)}_{S_3=1}}^2-2\big((z_1z_2)^3+(z_1z_3)^3+(z_2z_3)^3\big)$

$(z_1z_2)^3+(z_1z_3)^3+(z_2z_3)^3=$

$=\underbrace{(z_1z_2+z_1z_3+z_2z_3)}_{z_1z_2z_3}\big(\underbrace{(z_1z_2)^2+(z_1z_3)^2+(z_2z_3)^2}_{(z_1z_2z_3)^2-2z_1z_2z_3}-\underbrace{(z_1z_2+z_1z_3+z_2z_3)}_{z_1z_2z_3}\big)+3(z_1z_2z_3)^2$

$=z_1z_2z_3\big((z_1z_2z_3)^2-3z_1z_2z_3\big)+3(z_1z_2z_3)^2=(z_1z_2z_3)^2(z_1z_2z_3-3)+(z_1z_2z_3)^2\cdot3=(z_1z_2z_3)^3$

$\mathbf{S_6=1-2(z_1z_2z_3)^3}$

Let's take a look on $S_k$ for $k$ even.

$ \begin{cases} S_0=1+2(z_1z_2z_3)^0 \\ S_2=1-2(z_1z_2z_3)^1 \\ S_4=1+2(z_1z_2z_3)^2 \\ S_6=1-2(z_1z_2z_3)^3 \end{cases} $

And for $k$ odd we have $S_1=S_3=1$.

I don't think it's a coincidence.

My hypothesis:

$ S_k= \begin{cases} 1+2(-z_1z_2z_3)^\frac{k}{2}&k\text{ even} \\ 1&k\text{ odd} \end{cases} $

So the answer would be $S_{2013}=1$ which, judging by the answers that are given at the back of the textbook, is correct.

The question is, why is that? Can anyone prove it by any means? I thought of induction but I couldn't solve it.

EDIT:

I've just calculated $S_8$ and it is $1+2(z_1z_2z_3)^4$.

Also $(z_1z_2)^4+(z_1z_3)^4+(z_2z_3)^4=(z_1z_2z_3)^4+2(z_1z_2z_3)^2$.

Answer 1

Consider the cubic polynmomial with $z_1,z_2,z_3$ as its only roots, id est the polynomial $$(x-z_1)(x-z_2)(x-z_3)=x^3-(z_1+z_2+z_3)x^2+(z_1z_2+z_2z_3+z_1z_3)x-z_1z_2z_3=x^3-x^2+z_1z_2z_3x-z_1z_2z_3$$ using the fact that you proved above $z_1z_2+z_2z_3+z_1z_3=z_1z_2z_3$. For the sake of simplicity we assume $z_1z_2z_3=a$. then our equation is $x^3-x^2+ax-a=0$ which is satisfied by $z_1,z_2,z_3$, so they must also satisfy $x^m=x^{m-1}-ax^{m-2}+ax^{m-3}$, on adding these equations for $z_1,z_2,z_3$, we get that $$S_m=S_{m-1}-aS_{m-2}+aS_{m-3} \implies S_m-S_{m-1}=-a(S_{m-2}-S_{m-3})\implies \frac{S_m-S_{m-1}}{S_{m-2}-S_{m-3}}=-a$$ Let's call this equation $f(m)$. For $m$ even, multiplying equations $f(m),f(m-2),\cdots , f(4)$, we get $$\frac{S_m-S_{m-1}}{S_2-S_1}=(-a)^{\frac{m}{2}-1}\implies S_{m}-S_{m-1}=(-a)^{\frac{m}{2}-1}(-2a)=2(-a)^{\frac{m}{2}}$$ using the values of $S_2$ calculated by you. Similarly for odd $m$, $$S_m-S_{m-1}=-2(-a)^{\frac{m-1}{2}}$$ therefore for even $m$, we write $$S_{m}-S_{m-1}=2(-a)^{\frac{m}{2}}$$ $$S_{m-1}-S_{m-2}=-2(-a)^{\frac{m-2}{2}}$$ upto $$S_2-S_1=2(-a)$$ Add all these up and we get $$S_m-S_1=2(-a)^{\frac{m}{2}}\implies S_m=1+2(-a)^{\frac{m}{2}}$$ which is what you have hypothesized. Similar work will let you to the value of $S_m$ for odd $m$ which is indeed $1$.

Answer 2

You are in the right direction by finding the equality $z_1z_2 +z_2z_3+z_3z_1 = z_1z_2z_3$. Let denote it be $t$.

Hence, $z_{1,2,3}$ are 3 roots of this cubic equation $$x^3-x^2+tx-t =0 \tag{1}$$

Denote $S_k=z_1^k+z_2^k+z_3^k$, from $(1)$ you have $$S_n-S_{n-1}+tS_{n-2}-tS_{n-3} = 0 \tag{2}$$

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With $(2)$, you can check easily your hypothesis by induction.

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PS: A general solution can also be solved with this following method:

As the 3 roots of $(1)$ is $1, \pm i\sqrt{t}$, the formula of $S_n$ can be expressed as

$$S_n = a_1(i\sqrt{t})^n+a_2(-i\sqrt{t})^n+a_3 \tag{3}$$ with $a_i \in \Bbb C$ and is the solution of a system of linear equations $$ \begin{cases} a_1+a_2+a_3 = S_0 =3\\ i\sqrt{t}a_1-i\sqrt{t}a_2+a_3 = S_1 =1\\ -ta_1-ta_2+a_3 = S_2 =1+2t\\ \end{cases} \tag{4} $$

Solving $(4)$ and replacing the roots to $(3)$, you obtain the closed form expression of $S_n$ (in particular, $S_{2003}$) in function of $t = z_1z_2z_3$.