How do you find the fourth roots of $625 i$ $625i$ ?

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How do you find the fourth roots of $625 i$ $625i$ ?

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Answer 1 · 2016-09-07T03:41:55

Using $c i s θ = cos θ + i sin θ$ $cis theta = cos theta + i sin theta$ , the roots are

$5 c i s (\frac{π}{8}), 5 c i s (\frac{5}{8} π), 5 c i s (\frac{9}{8} π) and 5 c i s (\frac{13}{8} π)$ $5cis(pi/8), 5 cis(5/8pi), 5 cis(9/8pi) and 5 cis(13/8pi)$
.
= $5 (\pm \frac{\sqrt{2 + \sqrt{2}}}{2} \pm i \frac{\sqrt{2 - \sqrt{2}}}{2})$ $5(+-sqrt(2+sqrt 2)/2+-isqrt(2-sqrt 2)/2)$

Explanation:

Using $c i s θ$ $cis theta$ for cos theta +i sin theta#,

$c i s ((2 k π + \frac{π}{2}) i) = c i s (\frac{π}{2} i) = i$ $cis((2kpi+pi/2)i)=cis(pi/2i)=i$ , for k=0, +_1, +-2, +-3, ...

So, ${(625 i)}^{\frac{1}{4}} = 625^{\frac{1}{4}} {(c i s ((2 k π + \frac{π}{2}) i) = c i s (\frac{π}{2} i))}^{\frac{1}{4}}$ $(625i)^(1/4)=625^(1/4)( cis((2kpi+pi/2)i)=cis(pi/2i))^(1/4)$

$= 5 (c i s (\frac{2 k π + \frac{π}{2}}{4}) i), k = 0, 1, 2, 3$ $=5( cis((2kpi+pi/2)/4)i), k =0, 1, 2, 3$ , omitting other k values that

produce repetitions. So, the root are

$= 5 c i s (\frac{π}{8}), 5 c i s (\frac{5}{8} π), 5 c i s (\frac{9}{8} π) and 5 c i s (\frac{13}{8} π)$ $=5cis(pi/8), 5 cis(5/8pi), 5 cis(9/8pi) and 5 cis(13/8pi)$ .

Using $cos (\frac{π}{8}) = \sqrt{\frac{1 + cos (\frac{π}{4})}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $cos (pi/8)=sqrt((1+cos(pi/4))/2)=sqrt(2+sqrt 2)/2$ and,

likewise,

$sin (\frac{π}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $sin(pi/8)= sqrt(2-sqrt 2)/2$ , the answer can be presented as

$5 (\pm \frac{\sqrt{2 + \sqrt{2}}}{2} \pm i \frac{\sqrt{2 - \sqrt{2}}}{2})$ $5(+-sqrt(2+sqrt 2)/2+-isqrt(2-sqrt 2)/2)$

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How do you find the fourth roots of $625 i$ $625i$ ?

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How do you find the fourth roots of $625 i$ $625i$ ?