Question 1167604

Consider {{{z^5-i=0}}}

By finding the roots in {{{cis(theta)}}} form, and using appropriate substitutions,

Show:

{{{(z-i)(z^2-(2isin(pi/10))z-1)(z^2+(2isin(3pi/10))z-1)=0}}}

 {{{z^5-i=0}}}
 {{{i=z^5}}} => true if {{{z=i}}} ...(recall that {{{i^5=i}}})


substitute {{{z=i}}}

{{{(z-i)(z^2-(2*i*sin(pi/10))z-1)(z^2+(2i*sin(3pi/10))z-1)=0}}}

{{{(z-z) (z^2-(2z*sin(pi/10))z-1) (z^2+(2z*sin(3pi/10))z-1)=0}}}

{{{(0)(z^2-(1/2) (sqrt(5) - 1) z^2-1) (z^2+(1/2) (1 + sqrt(5)) z^2-1)=0}}}..........factored form, first factor is {{{0}}}, so whole product is{{{ 0}}}