If z=√32+ i2 (i=-1), then (1+iz+z^5+iz^8)^9 is equal to:

Question

If $z=\dfrac{\sqrt3}2+\dfrac i2$ ($i=\sqrt{-1}$), then $(1+iz+z^5+iz^8)^9$ is equal to:

Accepted Answer

$z=e^{i\pi/6}=-i\omega$ where $\omega$ is a cube root of unity. Then

$$1+iz+z^5+iz^8=1+\omega-i\omega^2+i\omega^2=1+\omega,$$

$$(1+iz+z^5+iz^8)^9=(1+\omega)^9=(-\omega^2)^9=-\omega^{18}=-1.$$

Solutions · 1