How do you find the sixth roots of $64 i$ $64i$ ?

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How do you find the sixth roots of $64 i$ $64i$ ?

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Answer 1 · 2016-09-26T06:31:01

The sixth roots are:

$\pm (\frac{1}{2} (\sqrt{6} + \sqrt{2}) + \frac{1}{2} (\sqrt{6} - \sqrt{2}) i)$ $+-(1/2(sqrt(6)+sqrt(2)) + 1/2(sqrt(6)-sqrt(2))i)$

$\pm (\frac{1}{2} (\sqrt{6} - \sqrt{2}) + \frac{1}{2} (\sqrt{6} + \sqrt{2}) i)$ $+-(1/2(sqrt(6)-sqrt(2)) + 1/2(sqrt(6)+sqrt(2))i)$

$\pm (- \sqrt{2} + \sqrt{2} i)$ $+-(-sqrt(2) + sqrt(2)i)$

Explanation:

De Moivre's formula tells us that:

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(cos theta + i sin theta)^n = cos n theta + i sin n theta$

Hence if $θ = \frac{π}{12}$ $theta = pi/12$ we find:

${(cos (\frac{π}{12}) + i sin (\frac{π}{12}))}^{6} = cos (\frac{π}{2}) + i sin (\frac{π}{2}) = 0 + i (1) = i$ $(cos (pi/12) + i sin (pi/12))^6 = cos (pi/2) + i sin (pi/2) = 0 + i (1) = i$

So $cos (\frac{π}{12}) = i sin (\frac{π}{12})$ $cos (pi/12) = i sin (pi/12)$ is a sixth root of $i$ $i$ .

We can form other sixth roots by adding any multiple of $\frac{2 π}{6} = \frac{π}{3}$ $(2pi)/6 = pi/3$ to theta. So in general we find:

${(cos (\frac{1 + 4 k}{12} π) + i sin (\frac{1 + 4 k}{12} π))}^{6} = cos (\frac{π}{2} + 2 k π) + i sin (\frac{π}{2} + 2 k π) = i$ $(cos ((1+4k)/12 pi) + i sin ((1+4k)/12 pi))^6 = cos (pi/2 + 2kpi) + i sin (pi/2 + 2kpi) = i$

Also we have:

${(r (cos θ + i sin θ))}^{n} = r^{n} (cos n θ + i sin θ)$ $(r(cos theta + i sin theta))^n = r^n (cos n theta + i sin theta)$

$2^{6} = 64$ $2^6 = 64$

Hence the sixth roots of $64 i$ $64i$ are:

$2 (cos (\frac{1 + 4 k}{12} π) + i sin (\frac{1 + 4 k}{12} π))$ $2(cos ((1+4k)/12 pi) + i sin ((1+4k)/12 pi))" "$ for $k = 0, 1, 2, 3, 4, 5$ $k = 0, 1, 2, 3, 4, 5$

You are probably aware that:

$sin (\frac{π}{4}) = cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin (pi/4) = cos (pi/4) = sqrt(2)/2$

$sin (\frac{π}{6}) = cos (\frac{π}{3}) = \frac{1}{2}$ $sin (pi/6) = cos (pi/3) = 1/2$

$cos (\frac{π}{6}) = sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $cos (pi/6) = sin (pi/3) = sqrt(3)/2$

$sin (α - β) = sin α cos β - sin β cos α$ $sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha$

$cos (α - β) = cos α cos β + sin α sin β$ $cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta$

So we find:

$sin (\frac{π}{12}) = sin (\frac{π}{3} - \frac{π}{4})$ $sin (pi/12) = sin (pi/3 - pi/4)$

$sin (\frac{π}{12}) = sin (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{4}) cos (\frac{π}{3})$ $color(white)(sin (pi/12)) = sin (pi/3) cos (pi/4) - sin (pi/4) cos (pi/3)$

$sin (\frac{π}{12}) = (\frac{\sqrt{3}}{2}) (\frac{\sqrt{2}}{2}) - (\frac{\sqrt{2}}{2}) (\frac{1}{2})$ $color(white)(sin (pi/12)) = (sqrt(3)/2) (sqrt(2)/2) - (sqrt(2)/2) (1/2)$

$sin (\frac{π}{12}) = \frac{1}{4} (\sqrt{6} - \sqrt{2})$ $color(white)(sin (pi/12)) = 1/4(sqrt(6)-sqrt(2))$

$cos (\frac{π}{12}) = cos (\frac{π}{3} - \frac{π}{4})$ $cos (pi/12) = cos (pi/3 - pi/4)$

$cos (\frac{π}{12}) = cos (\frac{π}{3}) cos (\frac{π}{4}) + sin (\frac{π}{3}) sin (\frac{π}{4})$ $color(white)(cos (pi/12)) = cos (pi/3) cos (pi/4) + sin (pi/3) sin (pi/4)$

$cos (\frac{π}{12}) = (\frac{1}{2}) (\frac{\sqrt{2}}{2}) + (\frac{\sqrt{3}}{2}) (\frac{\sqrt{2}}{2})$ $color(white)(cos (pi/12)) = (1/2) (sqrt(2)/2) + (sqrt(3)/2) (sqrt(2)/2)$

$cos (\frac{π}{12}) = \frac{1}{4} (\sqrt{6} + \sqrt{2})$ $color(white)(cos (pi/12)) = 1/4(sqrt(6)+sqrt(2))$

We also have:

$sin (\frac{5 π}{12}) = sin (\frac{π}{2} - \frac{π}{12}) = cos (\frac{π}{12}) = \frac{1}{4} (\sqrt{6} + \sqrt{2})$ $sin ((5pi)/12) = sin (pi/2 - pi/12) = cos (pi/12) = 1/4(sqrt(6)+sqrt(2))$

$cos (\frac{5 π}{12}) = cos (\frac{π}{2} - \frac{π}{12}) = sin (\frac{π}{12}) = \frac{1}{4} (\sqrt{6} - \sqrt{2})$ $cos ((5pi)/12) = cos (pi/2 - pi/12) = sin (pi/12) = 1/4(sqrt(6)-sqrt(2))$

So all the sixth roots are:

$\pm 2 (cos (\frac{π}{12}) + i sin (\frac{π}{12})) = \pm (\frac{1}{2} (\sqrt{6} + \sqrt{2}) + \frac{1}{2} (\sqrt{6} - \sqrt{2}) i)$ $+-2(cos (pi/12) + i sin (pi/12)) = +-(1/2(sqrt(6)+sqrt(2)) + 1/2(sqrt(6)-sqrt(2))i)$

$\pm 2 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12})) = \pm (\frac{1}{2} (\sqrt{6} - \sqrt{2}) + \frac{1}{2} (\sqrt{6} + \sqrt{2}) i)$ $+-2(cos ((5pi)/12) + i sin ((5pi)/12)) = +-(1/2(sqrt(6)-sqrt(2)) + 1/2(sqrt(6)+sqrt(2))i)$

$\pm 2 (cos (\frac{9 π}{12}) + i sin (\frac{9 π}{12})) = \pm (- \sqrt{2} + \sqrt{2} i)$ $+-2(cos ((9pi)/12) + i sin ((9pi)/12)) = +-(-sqrt(2) + sqrt(2)i)$

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How do you find the sixth roots of $64 i$ $64i$ ?

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