If $z=\dfrac12+\dfrac{\sqrt{3}}{2}i$, find the value of $z^4+z^9$ [duplicate]
As you have found that $z=e^{\pi i/3}$ then $z^3=-1$ and thus $$ z^4+z^9= z(z^3)+(z^3)^3= -z-1=-\frac{3}{2}-\frac{\sqrt{3}}{2}i $$
We can use the the exponential form: $$z^4=\left(\dfrac12+\dfrac{\sqrt{3}}{2}i\right)^4=e^{4\cdot\frac{\pi}{3}i}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$$ And: $$z^9=\left(\dfrac12+\dfrac{\sqrt{3}}{2}i\right)^9=e^{9\cdot\frac{\pi}{3}}=e^{3\pi}=e^{\pi}=-1$$ So, the value we wanted to compute is: $$-\frac{3}{2}-\frac{\sqrt{3}}{2}i$$ Your procedure is right. You have only to express $z^4$ and $z^9$ in algebric form and then sum together.