Help With Complex Plane

Question

Hello everyone I have a number w that equals to:

w = $-0.5 + 0.5 \cdot \sqrt{3} \cdot i$

And I have to proof that:

$(w-1)^{n} + (\bar{w}-1)^{n} \in$$ \{{0 , \pm \sqrt{3^n} , \pm \sqrt{3^{n+1}} , \pm 2 \sqrt{3^n} }\}$

Answer 1

We have that

$$ w - 1 = - \frac{3} {2} + \frac{{i\sqrt 3 }} {2} $$ thus $ r = \left| {w - 1} \right| = \sqrt {a^2 + b^2 } = \sqrt 3 $ and $$ \left\{ \begin{gathered} \cos \theta = \frac{a} {r} = - \frac{3} {{2\sqrt 3 }} = - \frac{{\sqrt 3 }} {2} \hfill \\ \sin \theta = \frac{b} {r} = \frac{1} {2} \hfill \\ \end{gathered} \right. $$ therefore $\theta=\frac{5\pi}{6}$ Hence $$ w - 1 = \left( {\sqrt 3 } \right)\left( {\cos \frac{{5\pi }} {6} + i\sin \frac{{5\pi }} {6}} \right) $$ and from De Moivre's Theorem you have $$ \left( {w - 1} \right)^n = \left( {\sqrt 3 } \right)^n \left( {\cos \frac{{5\pi n}} {6} + i\sin \frac{{5\pi n}} {6}} \right) .$$ In the same way $$ \begin{gathered} \overline w - 1 = - \frac{3} {2} - i\frac{{\sqrt 3 }} {2} \hfill \\ r = \sqrt 3 \hfill \\ \cos \theta = - \frac{{\sqrt 3 }} {2} \hfill \\ \sin \theta = - \frac{1} {2} \hfill \\ \theta = \frac{{7\pi }} {6} = - \frac{{5\pi }} {6} \hfill \\ \end{gathered} $$ thus $$ \left( {\overline w - 1} \right)^n = \left( {\sqrt 3 } \right)^n \left( {\cos \frac{{5\pi n}} {6} - i\sin \frac{{5\pi n}} {6}} \right) $$ Now $$ \left( {w - 1} \right)^n + \left( {\overline w - 1} \right)^n = 2\left( {\sqrt 3 } \right)^n \left( {\cos \frac{{5\pi n}} {6}} \right) $$ Now, the sequence $$ a_n = \cos \left( {\frac{{5\pi n}} {6}} \right) $$ is periodic with period $12$. Its values are

1. $a_1=-\sqrt{3}/2$
2. $a_2=1/2$
3. $a_3=0$
4. $a_4=-1/2$
5. $a_5=\sqrt{3}/2$
6. $a_6=-1$
7. $a_7=\sqrt{3}/2$
8. $a_8=-1/2$
9. $a_9=0$
10. $a_{10}=1/2$
11. $a_{11}=-\sqrt{3}/2$
12. $a_{12}=1$.

After that, the values are repeating with period $12$. If you combine this values with the factor $2(\sqrt{3})^n$ you get the result.