Question 978131
If the complex number {{{r(cos(theta)+i*sin(theta))}}} ,
with {{{r>=0}}} and {{{0^o<=theta<360^o}}} ,
is the complex cube root of {{{343(cos(291^o)+i*sin(291^o))}}} , then
{{{(r(cos(theta)+i*sin(theta)))^3=r^3(cos(3theta)+i*sin(3theta))=343(cos(291^o+n*360^o)+i*sin(291^o+n*360^o))}}} for {{{system(n=0,n=1,"or",n=2)}}} .
For that it must be
{{{r^3=343}}}--->{{{r=7}}} and
{{{3theta=291^o+n*360^o}}} for {{{system(n=0,n=1,"or",n=2)}}}
Dividing both sides of the equation by {{{3}}} we get
{{{3theta/3=(291^o+n*360^o)/3}}}
{{{theta=291^o/3+n*360^o/3}}}
{{{theta=97^o+n*120^o}}}--->{{{system(theta=97^o+0*120^o=97^o+0^o=97^o,theta=97^o+1*120^o=97^o+120^o=217^o,"or",theta=97^o+2*120^o=97^o+240^o=337^o)}}}