Question 1183762
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The answer from the other tutor is incomplete/incorrect.  There are 6 6th roots of 1 (or any other number); there are 5 5th roots of 32 (or any other number).  The response from the other tutor only shows the real roots -- 2 of the 6 6th roots of 1, and 1 of the 5 5th roots of 32.<br>
De Moivre's Theorem solves this kind of problem easily.<br>
Given a number represented in trigonometric form<br>
{{{N=a*cis(theta)}}} or {{{N=a*e^(i*theta)}}},<br>
de Moivre's Theorem says that the "primary" n-th root is<br>
{{{a^(1/n)*cis(theta/n)}}}<br>
and the other n-th roots have the same magnitude as the primary root and are distributed around the Argand plane at intervals of {{{2pi/n}}}<br>
(a) The 6 6th roots of 1<br>
{{{1 = 1*cis(0)}}}<br>
The primary root is<br>
{{{1^(1/6)*cis(0/6) = 1*cis(0) = 1}}}<br>
The other 6th roots of 1 are distributed about the Argand plane at intervals of {{{2pi/6=pi/3}}}; so the 6 6th roots of 1 are<br>
cis(0) = 1
cis(pi/3) = 1/2+i*sqrt(3)/2
cis(2pi/3) = -1/2+i*sqrt(3)/2
cis(pi) = -1
cis(4pi/3) = -1/2-i*sqrt(3)/2
cis(5pi/3) = 1/2-i*sqrt(3)/2<br>
(b) The 5 5th roots of 32<br>
The primary root is<br>
{{{32^(1/5)*cis(0/5) = 2*cis(0) = 2}}}<br>
The other 5th roots of 1 are distributed about the Argand plane at intervals of 2*cis(pi/5); so the 5 5th roots of 32 are:<br>
2*cis(0) = 2
2*cis(pi/5)
2*cis(2pi/5)
2*cis(3pi/5)
2*cis(4pi/5)<br>