2011/036) Find z^2010 where The complex number z is such that z^9 = zz', where z and z' are complex conjugates.

let z = r e^(it)

then z^9 = r^9 e^(9it)

zz' = r^2 = r^9 e^(9it)

so e^(9it) = r^7 or r = 1 and hence zz' = 1

so z^9 = 1 so z^2007 = 1 and so z^2010 = z^3 = cube root of 1

z = r = 0 is also another solution

so solutions are 3 cube roots of 1 and zero