# Solve for n, (-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2 ?

Mar 12, 2017

$n = \frac{3 m}{4}$, where $m$ is an integer

#### Explanation:

We have to evaluate ${\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)}^{3} + {\left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}^{2 n}$

As $\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$ and

$\left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)$

using De Moivre's Theorem

${\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)}^{3} = \cos \left(\frac{2 \pi}{3} \times 3\right) + i \sin \left(\frac{2 \pi}{3} \times 3\right)$

= $\cos \left(2 \pi\right) + i \sin \left(2 \pi\right) = 1$

and ${\left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}^{2 n} = \cos \left(\frac{4 \pi}{3} \times 2 n\right) + i \sin \left(\frac{4 \pi}{3} \times 2 n\right)$

= $\cos \left(\frac{8 n \pi}{3}\right) + i \sin \left(\frac{8 n \pi}{3}\right)$

Therefore ${\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right)}^{3} + {\left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}^{2 n} = 2$

$\cos \left(\frac{8 n \pi}{3}\right) + i \sin \left(\frac{8 n \pi}{3}\right) = 1 = \cos \left(2 m \pi\right) + i \sin \left(2 m \pi\right)$,

where $m$ is an integer

i.e. $\frac{8 n \pi}{3} = 2 m \pi$ or $m = \frac{4 n}{3}$

and $n = \frac{3 m}{4}$, where $m$ is an integer

Mar 12, 2017

$n = \frac{3}{2} k$ with $k = 1 , 2 , 3 , \cdots$

#### Explanation:

${\left(- \frac{1}{2} + \iota \frac{\sqrt{3}}{2}\right)}^{3} + {\left(- \frac{1}{2} - \iota \frac{\sqrt{3}}{2}\right)}^{2 n} = 2$

We will using the complex number exponential representation

$x + i y = \rho {e}^{i \phi}$ where $\rho = \sqrt{{x}^{2} + {y}^{2}}$ and $\phi = \arctan \left(\frac{y}{x}\right)$

So $\rho = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = 1$ and $\phi = \arctan \left(\sqrt{3}\right) = \frac{\pi}{3}$

then we have

${e}^{i 3 \phi} + {e}^{- i 2 n \phi} = 2$

or

${e}^{i 3 \phi} + {e}^{- i 3 \phi \left(\frac{2 n}{3}\right)} = 2$

but ${e}^{i 3 \phi} = 1$ and ${e}^{- i 3 \phi} = 1$ so with $\frac{2 n}{3} = k , k = 1 , 2 , 3 , \cdots$

$1 + {1}^{k} = 2$

then

$n = \frac{3}{2} k$ with $k = 1 , 2 , 3 , \cdots$