Solve for $n$ $n$ , ${(- \frac{1}{2} + ι \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - ι \frac{\sqrt{3}}{2})}^{2 n} = 2$ $(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2$ ?

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Solve for $n$ $n$ , ${(- \frac{1}{2} + ι \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - ι \frac{\sqrt{3}}{2})}^{2 n} = 2$ $(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2$ ?

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Answer 1 · 2017-03-12T16:29:54

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

Explanation:

We have to evaluate ${(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - i \frac{\sqrt{3}}{2})}^{2 n}$ $(-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)$

As $(- \frac{1}{2} + i \frac{\sqrt{3}}{2}) = cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})$ $(-1/2+isqrt3/2)=cos((2pi)/3)+isin((2pi)/3)$ and

$(- \frac{1}{2} - i \frac{\sqrt{3}}{2}) = cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3})$ $(-1/2-isqrt3/2)=cos((4pi)/3)+isin((4pi)/3)$

using De Moivre's Theorem

${(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{3} = cos (\frac{2 π}{3} \times 3) + i sin (\frac{2 π}{3} \times 3)$ $(-1/2+isqrt3/2)^3=cos((2pi)/3xx3)+isin((2pi)/3xx3)$

= $cos (2 π) + i sin (2 π) = 1$ $cos(2pi)+isin(2pi)=1$

and ${(- \frac{1}{2} - i \frac{\sqrt{3}}{2})}^{2 n} = cos (\frac{4 π}{3} \times 2 n) + i sin (\frac{4 π}{3} \times 2 n)$ $(-1/2-isqrt3/2)^(2n)=cos((4pi)/3xx2n)+isin((4pi)/3xx2n)$

= $cos (\frac{8 n π}{3}) + i sin (\frac{8 n π}{3})$ $cos((8npi)/3)+isin((8npi)/3)$

Therefore ${(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - i \frac{\sqrt{3}}{2})}^{2 n} = 2$ $(-1/2+isqrt3/2)^3+(-1/2-isqrt3/2)^(2n)=2$

$cos (\frac{8 n π}{3}) + i sin (\frac{8 n π}{3}) = 1 = cos (2 m π) + i sin (2 m π)$ $cos((8npi)/3)+isin((8npi)/3)=1=cos(2mpi)+isin(2mpi)$ ,

where $m$ $m$ is an integer

i.e. $\frac{8 n π}{3} = 2 m π$ $(8npi)/3=2mpi$ or $m = \frac{4 n}{3}$ $m=(4n)/3$

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

Answer 2 · 2017-03-12T17:14:15

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

Explanation:

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

We will using the complex number exponential representation

$x + i y = ρ e^{i ϕ}$ $x+iy=rho e^(i phi)$ where $ρ = \sqrt{x^{2} + y^{2}}$ $rho=sqrt(x^2+y^2)$ and $ϕ = arctan (\frac{y}{x})$ $phi=arctan(y/x)$

So $ρ = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = 1$ $rho = sqrt((1/2)^2+(sqrt3 /2)^2) = 1$ and $ϕ = arctan (\sqrt{3}) = \frac{π}{3}$ $phi=arctan(sqrt(3))=pi/3$

then we have

$e^{i 3 ϕ} + e^{- i 2 n ϕ} = 2$ $e^(i 3 phi)+e^(-i 2n phi)=2$

or

$e^{i 3 ϕ} + e^{- i 3 ϕ (\frac{2 n}{3})} = 2$ $e^(i 3 phi)+e^(-i3phi((2n)/3))=2$

but $e^{i 3 ϕ} = 1$ $e^(i3phi)= 1$ and $e^{- i 3 ϕ} = 1$ $e^(-i3phi)= 1$ so with $\frac{2 n}{3} = k, k = 1, 2, 3, \dots$ $(2n)/3=k, k=1,2,3,cdots$

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

then

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

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Solve for $n$ $n$ , ${(- \frac{1}{2} + ι \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - ι \frac{\sqrt{3}}{2})}^{2 n} = 2$ $(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2$ ?

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Explanation:

Explanation:

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Solve for $n$ $n$ , ${(- \frac{1}{2} + ι \frac{\sqrt{3}}{2})}^{3} + {(- \frac{1}{2} - ι \frac{\sqrt{3}}{2})}^{2 n} = 2$ $(-1/2+ iota sqrt(3)/2)^3 + (-1/2-iota sqrt(3)/2)^(2n)=2$ ?