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\displaymath \frac{z_{n+1}-z_n}{z_n-z_{n-1}} = d e^{i\theta}
\mathbf{a}_1 + \mathbf{a}_2 + \dots + \mathbf{a}_n = \mathbf{0}. \\[br]\therefore \mathbf{a}_1 \cdot [\mathbf{a}_1 + \mathbf{a}_2 + \dots + \mathbf{a}_n] = 0 \\[br]\therefore \alpha^2 + \underbrace{\beta + \beta + \dots + \beta}_{n+1} = 0 \\ [br](n-1)\beta = -\alpha^2.
\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} ; \mathbf{a}_1 \cdot \mathbf{a}_1 = \alpha^2 = 1. \\[br]\text{Let } \mathbf{a}_i = \begin{pmatrix} x_i \\ y_i \end{pmatrix}. \\[br]\mathbf{a}_i \cdot \mathbf{a}_i = |\mathbf{a}_i | = 1 \text{ for all } 0 < i \leq n \Rightarrow \text{ vectors }\mathbf{a}_i \text{ lie on a circle about O} . \\[br]\mathbf{a}_1 \cdot \mathbf{a}_i = \mathbf{a}_1 \cdot \mathbf{a}_j \Rightarrow x_i = x_j \text{ for all } 0 < i \leq j \leq n. \\[br]\therefore n = 2 \text{ or } 3. \\[br]n = 2 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1 \\ 0 \end{pmatrix} \\[br]n = 3 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1/2 \\ \sqrt{3} /2 \end{pmatrix} , \mathbf{a}_3 = \begin{pmatrix} -1/2 \\ -\sqrt{3} /2 \end{pmatrix} \text{ or vice-versa} .
\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \Rightarrow \alpha^2 = 1, x_i = x_j \text{ for all } 1 < i \leq j \leq n. \\[br]n \leq 4. \\[br]n = 2 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} \\[br]n = 3 \Rightarrow \mathbf{a}_2 = \begin{pmatrix} -1/2 \\ \sqrt{3} /2 \\ 0 \end{pmatrix} , \mathbf{a}_3 = \begin{pmatrix} -1/2 \\ -\sqrt{3} /2 \\ 0 \end{pmatrix} \text{ or vice-versa} .\\[br]n = 4 \Rightarrow \text{ vertices of regular tetrahedron} .\\[br]\beta = -1/3. \\[br]\mathbf{a}_2 \cdot \mathbf{a}_1 = -1/3 \Rightarrow x_2 = -1/3. \\[br]a_2 = \frac{1}{9} + y_2^2 = 1 \Rightarrow y_2 = \pm 2\sqrt{2} /3. \\[br]\text{If } y_2 = 2\sqrt{2} /3 \text{ then, by considering } \mathbf{a}_2 \cdot \mathbf{a}_3 , \text{ it is obvious that } y_3 = -\sqrt{2} /3 \text{; similarly, } y_4 = -\sqrt{2} /3. \\[br]\text{By considering } |\mathbf{a}_3 |, \text{ we see that } z_3 = \pm \sqrt{6} /3. \text{If } z_3 = +\sqrt{6} /3, \text{ then } z_4 = -\sqrt{6} /3. \\[br]\therefore \mathbf{a}_2 = \dfrac{1}{3}\begin{pmatrix} -1 \\ 2\sqrt{2} \\ 0 \end{pmatrix} , \mathbf{a}_3 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -\sqrt{2} \\ \sqrt{6} \end{pmatrix} , \mathbf{a}_4 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -\sqrt{2} \\ -\sqrt{6} \end{pmatrix} . \\[br]\text{Similarly, if } y_2 = -2\sqrt{2} /3 , \text{ then} \\[br]\mathbf{a}_2 = \dfrac{1}{3}\begin{pmatrix} -1 \\ -2\sqrt{2} \\ 0 \end{pmatrix} , \mathbf{a}_3 = \dfrac{1}{3}\begin{pmatrix} -1 \\ \sqrt{2} \\ \sqrt{6} \end{pmatrix} , \mathbf{a}_4 = \dfrac{1}{3}\begin{pmatrix} -1 \\ \sqrt{2} \\ -\sqrt{6} \end{pmatrix} .
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