For non-zero real $x$, $y$, $z$ and complex $a$, $b$, $c$ with $|a|=|b|=|c|$, if $x+y+z=0=ax+by+cz$, then $a=b=c$

Question

Let $x,y,z \in \Bbb R-\{0\}$ and $\alpha,\beta,\gamma \in \Bbb C$ such that $|\alpha|=|\beta|=|\gamma|=1$. If $x+y+z=0=\alpha x+\beta y+\gamma z$, then prove that $\alpha=\beta=\gamma$.

My approach:

From the given data, $\alpha=\cos\theta_1+i\sin\theta_1$ where $\theta_1 \in [0,2\pi)$. Similarly, $\beta=\cos\theta_2+i\sin\theta_2$ and $\gamma=\cos\theta_3+i\sin\theta_3$. where $\theta_2,\theta_3 \in [0,2\pi)$.

Then we get homogeneous system of equations as below, $$ \left\{ \begin{array}{c} x+y+z=0 \\ x\cos\theta_1+y\cos\theta_2+z\cos\theta_3=0 \\ x\sin\theta_1+y\sin\theta_2+z\sin\theta_3=0 \end{array} \right. $$

Then, $$ \begin{vmatrix} 1 & 1 & 1 \\ \cos\theta_1 & \cos\theta_2 & \cos\theta_3 \\ \sin\theta_1 & \sin\theta_2 & \sin\theta_3 \\ \end{vmatrix}=0$$ because $x,y,z$ are non-zero.

On solving the determinant we get,

$\sin(\theta_2-\theta_3)+\sin(\theta_3-\theta_1)+\sin(\theta_1-\theta_2)=0$

$\Rightarrow 2\sin(\frac {\theta_2-\theta_3+\theta_3-\theta_1}2)\cos(\frac {\theta_2-\theta_3-\theta_3+\theta_1}2)-\sin(\theta_2-\theta_1)=0$

$\Rightarrow 2\sin(\frac {\theta_2-\theta_1}2)\cos(\frac {\theta_1+\theta_2}2 -\theta_3) - 2\sin(\frac {\theta_2-\theta_1}2)\cos(\frac {\theta_2-\theta_1}2)=0$

$\Rightarrow \sin(\frac {\theta_2-\theta_1}2)(\cos(\frac {\theta_1+\theta_2}2 - \theta_3) - \cos(\frac {\theta_2-\theta_1}2))=0$

$\Rightarrow \sin(\frac {\theta_2-\theta_1}2)=0$ or $\cos(\frac {\theta_1+\theta_2}2 - \theta_3) - \cos(\frac {\theta_2-\theta_1}2)=0$

$\Rightarrow \theta_1=\theta_2$ or $\theta_1=\theta_3$ (because $\theta_1,\theta_2,\theta_3 \in [0,2\pi)).$

Hence $\alpha=\beta$ or $\alpha=\gamma$. :(

Answer 1

Assume that $\alpha = \beta \not= \gamma$.

We have $$(\gamma - \alpha)z =(\alpha)(x+y+z) + (\gamma - \alpha)z = \alpha x+ \beta y + \gamma z = 0$$

Since we assumed $\alpha \not= \gamma$, we have $z=0$, a contradiction.

Therefore, we have $\alpha = \beta = \gamma$, as desired. $\blacksquare$

Answer 2

$AB = 0$ does not exclude $A = B = 0$.

You can simply rearrange the equations:

$$ \begin{vmatrix} 1 & 1 & 1 \\ \cos\theta_2 & \cos\theta_3 & \cos\theta_1 \\ \sin\theta_2 & \sin\theta_3 & \sin\theta_1 \\ \end{vmatrix}=0$$

And use an analogous proof to show that ${\beta = \alpha} \vee {\beta = \gamma}$. And rearrange a final time to close the loop.

The only solution to $({\alpha = \beta} \vee {\alpha = \gamma}) \wedge ({\beta = \alpha} \vee {\beta = \gamma}) \wedge ({\gamma = \alpha} \vee {\gamma = \beta})$ is $\alpha = \beta = \gamma$.

Answer 3

Now give a different proof from above! First, we claim that: for any complex numbers $z_1,z_2$, we have $$|z_1+z_2|^2=|z_1|^2+|z_2|^2+2Re(z_1\overline{z_2}).$$ From
$$|\alpha|=|\beta|=|\gamma|=1$$ $$|x+y|^2=|-z|^2,$$ $$|\alpha x+\beta y|^2=|-\gamma z|^2.$$ Use above claim we can get: $$x^2+y^2+2xy=z^2,$$ $$x^2+y^2+2xyRe(\alpha\overline{\beta})=z^2,$$ so $Re(\alpha\overline{\beta})=1$ and $Im(\alpha\overline{\beta})=0$. (because $|\alpha|=|\beta|=1$)

We get $$\alpha=\beta.$$ In the same way, we can get $$\alpha=\beta=\gamma.$$

Answer 4

It is given that $x,y,z$ are non-zero real numbers. We also have $|\alpha|=|\beta|=|\gamma|=1$ and $x+y+z=0=\alpha x+\beta y+\gamma z$.

Now since $\alpha x+\beta y+\gamma z=0\implies \overline{\alpha}x+\overline{\beta}y+\overline{\gamma}z=0.$

We also have $\alpha x+\beta y=-\gamma z.$

Now since $\alpha\overline{\alpha}=|\alpha|^2=1, \beta\overline{\beta}=|\beta|^2=1$ and $\gamma\overline{\gamma}=|\gamma|^2=1$, we have $$\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=0$$ $$\implies \frac{x}{\alpha}+\frac{y}{\beta}=-\frac{z}{\gamma}...(2)$$

Multiplying $(1)$ and $(2)$ we have $$\left(\frac{x}{\alpha}+\frac{y}{\beta}\right)(\alpha x+\beta y)=z^2=(x+y)^2$$ $$\implies x^2+xy\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+y^2=x^2+2xy+y^2$$ $$\implies\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=2$$ $$\implies (\alpha-\beta)^2=0$$ $$\implies \alpha=\beta.$$ Similarly we will get $\beta=\gamma$. This implies that we have $\alpha=\beta=\gamma$. Hence proved.