Question 1181163
If 1 and w are 2 of the 5 roots of (w)^5 = 1, then prove the following:
a) (w)^2, (w)^3, and (w)^4 are the remaining roots of (w)^5 = 1
<pre>
where cis(&theta;) = cos(&theta;)+isin(&theta;),
The 5 fifth roots of 1 are 

{{{matrix(1,9,       cis(0^o)=1,",",cis(72^o),",",cis(144^o),",",cis(216^o),",",cis(288^o))}}}

{{{(cis(72^o)^"")^2=cis(2*72^o)=cis(144^o)}}}
{{{(cis(144^o)^"")^2=cis(2*144^o)=cis(288^o)}}}
{{{(cis(288^o)^"")^2=cis(2*288^o)=cis(576^o)=cis(576^o-360^o)=cis(216^o)}}}

{{{(cis(72^o)^"")^3=cis(3*72^o)=cis(216^o)}}}
{{{(cis(144^o)^"")^3=cis(3*144^o)=cis(432^o)=cis(432-360^o)=cis(72^o)}}}
{{{(cis(288^o)^"")^3=cis(3*288^o)=cis(864^o)=cis(864^o-2*360^o)=cis(144)}}}

{{{(cis(72^o)^"")^4=cis(4*72^o)=cis(288^o)}}}
{{{(cis(144^o)^"")^4=cis(4*144^o)=cis(576^o)=cis(576-360^o)=cis(216^o)}}}
{{{(cis(288^o)^"")^4=cis(4*288^o)=cis(1152^o)=cis(1152^o-3*360^o)=cis(72)}}}
</pre>b) 1 + w + (w)^2 + (w)^3 + (w)^4 = 0, provided w ≠ 1<pre>
{{{w^5=1}}}
{{{w^5-1=0}}}
{{{(w-1)(w^4+w^3+w^2+w+1)=0}}}
Since w ≠ 1, w-1 ≠ 0, so
{{{w^4+w^3+w^2+w+1 = 0}}}
{{{1+w+w^2+w^3+w^4=0}}}
</pre>c) (w)^5n = 1 for any integer n<pre>
{{{w^5 = 1}}}
Raise both sides to the 5th power:
{{{(w^5)^n=1^n}}}
{{{w^(5n)=1}}}

Edwin</pre>