Thursday, April 26, 2018

polynomials - Using the fifth roots of unity to find the roots of (z+1)5=(z1)5


The question I am working on starts of with:




Find the five fifth roots of unity and hence solve the following problems




I have done that and solved several questions using this, however when I came to the last question (the one in the title) I got stumped.



The five fifth root of unity are z=cos(2πk5)+isin(2πk5),kZ or more simply put w=cos(2π5)+isin(2π5) and the roots are z=1,w,w2,w3,w4.


Now, using this information we are supposed to find the roots of (z+1)5=(z1)5. However having tried some different approaches I don't know how to proceed. I tried simplifying it to 5z4+10z2+1=0 from which I suppose I can use the quadratic equation, but it does not utilize the five fifth roots of unity and so I wont get my answer in terms of w (which is the requirement).


Answer



Having calculated the fifth roots of unity 1,w,w2,w3,w4, i.e. the solutions to the equation u5=1, we can recast our equation as (z+1z1)5=1 provided z is not equal to 1.


It is easy to verify that z=1 is not a solution to the equation (z+1)5=(z1)5. So for z not equal to 1, we have that z+1z1=1,w,w2,w3,w4. Simple manipulation will allow you to express z in terms of these roots.


Note that setting z+1z1=1 will not yield a solution. This can be expected, since the equation we are solving is quartic and so has at most 4 distinct roots.


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