SOLUTION: 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

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 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      Log On


   

Question 1167604: Consider z%5E5-i=0
By finding the roots in cistheta form, and using appropriate substitutions,
Show:
%28z-i%29%28z%5E2-%282isin%28pi%2F10%29%29z-1%29%28z%5E2%2B%282isin%283pi%2F10%29%29z-1%29=0
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Consider z%5E5-i=0
By finding the roots in cis%28theta%29 form, and using appropriate substitutions,
Show:
%28z-i%29%28z%5E2-%282isin%28pi%2F10%29%29z-1%29%28z%5E2%2B%282isin%283pi%2F10%29%29z-1%29=0
z%5E5-i=0
i=z%5E5 => true if z=i ...(recall that i%5E5=i)

substitute z=i


..........factored form, first factor is 0, so whole product is+0

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
Consider z%5E5-i=0.
By finding the roots in cis%28theta%29 form, and using appropriate substitutions, show that
= 0.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 


Equation  z%5E5-i = 0  is the same as  z%5E5 = i.


One root is, obviously, z = i,  since  i%5E5 = i.


Let's list all the roots 

    z%5B1%5D = cis%28pi%2F%282%2A5%29%29 = cis%28pi%2F10%29,  

    z%5B2%5D = cis%28pi%2F10+%2B+2pi%2F5%29 = cis%28pi%2F10%2B4pi%2F10%29 = cis%285pi%2F10%29 = cis%28pi%2F2%29 = i, 

              (we just noticed it above !)

    z%5B3%5D = cis%28pi%2F10+%2B+4pi%2F5%29 = cis%28pi%2F10%2B8pi%2F10%29 = cis%289pi%2F10%29, 

    z%5B4%5D = cis%28pi%2F10+%2B+6pi%2F5%29 = cis%28pi%2F10%2B12pi%2F10%29 = cis%2813pi%2F10%29,

    z%5B5%5D = cis%28pi%2F10+%2B+8pi%2F5%29 = cis%28pi%2F10%2B16pi%2F10%29 = cis%2817pi%2F10%29.
 

Notice that  z%5B1%5D  and  z%5B3%5D  have opposite real parts and identical imaginary parts.    (*)
Similarly,   z%5B4%5D  and  z%5B5%5D  have opposite real parts and identical imaginary parts.    (**)


We can write the decomposition of  z%5E5-i  in the form of the product of linear binomials with the roots

    z%5E5-i = %28z-z%5B1%5D%29%2A%28z-z%5B2%5D%29%2A%28z-z%5B3%5D%29%2A%28z-z%5B4%5D%29%2A%28z-z%5B5%5D%29 =

                = .    (1)


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

    %28z-cis%28pi%2F10%29%29%2A%28z-cis%289pi%2F10%29%29 = z%5E2+-+%28cis%28pi%2F10%29%2Bcis%289%2F10%29%29+%2B+cis%28pi%2F10%29%2Acis%289pi%2F10%29.    (2)


Here  cis%28pi%2F10%29+%2B+cis%289pi%2F10%29 = 2isin%28pi%2F10%29,  as we noticed in (*),  and  cis%28pi%2F10%29%2Acis%289pi%2F10%29 = cis%2810pi%2F10%29 = cis%28pi%29 = -1.


Therefore, 

    %28z-cis%28pi%2F10%29%29%2A%28z-cis%289pi%2F10%29%29 = z%5E2+-+2isin%28pi%2F10%29+-1.



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

    %28z-cis%2813pi%2F10%29%29%2A%28z-cis%2817pi%2F10%29%29 = .    (3)


Here  cis%2813pi%2F10%29+%2B+cis%2817pi%2F10%29 = -2isin%283pi%2F10%29,  as we noticed in (**),  and  cis%2813pi%2F10%29%2Acis%2817pi%2F10%29 = cis%2830pi%2F10%29 = cis%283pi%29 = -1.


Therefore, 

    %28z-cis%2813pi%2F10%29%29%2A%28z-cis%2817pi%2F10%29%29 = z%5E2+%2B+2isin%283pi%2F10%29+-1.    (4)



Thus, combining everything in one piece, we get


    If  z%5E5-i = 0,  then  z%5E5-i = %28z-i%29%2A%28z%5E2+-+2isin%28pi%2F10%29+-1%29%2A%28z%5E2+%2B+2isin%283pi%2F10%29+-1%29 = 0.


QED.


At this point, the proof is complete.

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.