(z-1)power5=(z+1)power5?

Question

I foun four root of this theorem ,what is the fifth ?
z=_+(0.866+_0.5 i )

I know the z^5 canceled. but our teacher told me to find 5th root !

Answer 1

There is no 5th root as this is 4th order eqution as below

for complete solution please see

(z-1)^5 = (z+1)^5

Expanding both sides gives
z^5 - 5z^4 + 10z^3 - 10z^2 + 5z - 1
= z^5 + 5z^4 + 10z^3 + 10z^2 + 5z + 1
Therefore,
10z^4 + 20z^2 + 2 = 0
This is a quadratic equation in z^2 i.e we could say k = z^2 and get
10k^2 + 20k + 2 = 0
which gives
k = [-20 +or- sqrt(20^2 - 4.10.2)]/2.10
= [-20 +or- sqrt(320)]/20
= -1 +or- (2/5)(sqrt5)
and k is approximately -0.11 or -1.89
so, z^2 = -0.11 or -1.89
z = +or- 0.33i or +or- 1.38i
where 'i' is the complex number sqrt(-1)

Answer 2

There is no fifth root. When expanded, this equation becomes a 4th degree polynomial. z^5 cancels out.