Question 1167604
.
Consider {{{z^5-i=0}}}.

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

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



<pre>
Equation  {{{z^5-i}}} = 0  is the same as  {{{z^5}}} = i.


One root is, obviously, z = i,  since  {{{i^5}}} = i.


Let's list all the roots 

    {{{z[1]}}} = {{{cis(pi/(2*5))}}} = {{{cis(pi/10)}}},  

    {{{z[2]}}} = {{{cis(pi/10 + 2pi/5)}}} = {{{cis(pi/10+4pi/10)}}} = {{{cis(5pi/10)}}} = {{{cis(pi/2)}}} = i, 

              (we just noticed it above !)

    {{{z[3]}}} = {{{cis(pi/10 + 4pi/5)}}} = {{{cis(pi/10+8pi/10)}}} = {{{cis(9pi/10)}}}, 

    {{{z[4]}}} = {{{cis(pi/10 + 6pi/5)}}} = {{{cis(pi/10+12pi/10)}}} = {{{cis(13pi/10)}}},

    {{{z[5]}}} = {{{cis(pi/10 + 8pi/5)}}} = {{{cis(pi/10+16pi/10)}}} = {{{cis(17pi/10)}}}.
 

Notice that  {{{z[1]}}}  and  {{{z[3]}}}  have opposite real parts and identical imaginary parts.    (*)
Similarly,   {{{z[4]}}}  and  {{{z[5]}}}  have opposite real parts and identical imaginary parts.    (**)


We can write the decomposition of  {{{z^5-i}}}  in the form of the product of linear binomials with the roots

    {{{z^5-i}}} = {{{(z-z[1])*(z-z[2])*(z-z[3])*(z-z[4])*(z-z[5])}}} =

                = {{{(z-i)*(z-cis(pi/10))*(z-cis(9pi/10))*(z-cis(13pi/10))*(z-cis(17pi/10))}}}.    (1)


In this decomposition (1), second and third parentheses will give the product

    {{{(z-cis(pi/10))*(z-cis(9pi/10))}}} = {{{z^2 - (cis(pi/10)+cis(9/10)) + cis(pi/10)*cis(9pi/10)}}}.    (2)


Here  {{{cis(pi/10) + cis(9pi/10)}}} = {{{2isin(pi/10)}}},  as we noticed in (*),  and  {{{cis(pi/10)*cis(9pi/10)}}} = {{{cis(10pi/10)}}} = {{{cis(pi)}}} = -1.


Therefore, 

    {{{(z-cis(pi/10))*(z-cis(9pi/10))}}} = {{{z^2 - 2isin(pi/10) -1}}}.



Similarly, in decomposition (1), fourth and fifth parentheses will give the product

    {{{(z-cis(13pi/10))*(z-cis(17pi/10))}}} = {{{z^2 - (cis(13pi/10)+cis(17/10)) + cis(13pi/10)*cis(17pi/10)}}}.    (3)


Here  {{{cis(13pi/10) + cis(17pi/10)}}} = {{{-2isin(3pi/10)}}},  as we noticed in (**),  and  {{{cis(13pi/10)*cis(17pi/10)}}} = {{{cis(30pi/10)}}} = {{{cis(3pi)}}} = -1.


Therefore, 

    {{{(z-cis(13pi/10))*(z-cis(17pi/10))}}} = {{{z^2 + 2isin(3pi/10) -1}}}.    (4)



Thus, combining everything in one piece, we get


    If  {{{z^5-i}}} = 0,  then  {{{z^5-i}}} = {{{(z-i)*(z^2 - 2isin(pi/10) -1)*(z^2 + 2isin(3pi/10) -1)}}} = 0.


QED.


At this point, the proof is complete.
</pre>

Solved.



////////////////////////////



In her post, @MathLover1 incorrectly read the problem and incorrectly understood 
what the problem requested to prove.


So, her writing in her post is not a proof of the problem' statement 
and has nothing in common with what this problem requests to prove.


For the peace in your mind, simply ignore that post.