Is this right? Difference between evaluating and expressing to cartesian form $z= 1 + \sqrt{3} i $ Am i going right with this? 
$$z= 1 + \sqrt{3} i $$
i need to i) evaluate $z^9$ and ii) express in cartesian for $z^5$... Which i'm a bit confused with.
First what i did was find the polar form...
So height angle with opposite side $\sqrt{3}$ and adjacent 1 $$\theta = \frac{\pi}{3}$$
$$r=\sqrt{1^2+\sqrt{3}^2}$$ $$r=\sqrt{1+3}$$ $$r=2$$
Hence polar form is.... $$z=2e^{\frac{\pi}{3}i}$$
Then i did part i. Using de moivre's theorem..... $$z^9=\left(2e^{\frac{\pi}{3}i}\right)^{9}$$ $$z^9=512e^{3\pi i}$$
Now here's where i get confused... Does evaluate mean... i go beyond cartesian form?
$$z^9=512e^{3\pi i}$$ $$z^9=512(\cos{3\pi}+i\sin{3\pi})$$
Which equates to....
$$z^9=512(-1+i0)$$ $$z^9=-512$$
for part ii)
Again using de moivre's theorem $$z^5=\left(2e^{\frac{\pi}{3}i}\right)^{5}$$ $$z^5=32e^{\frac{5\pi}{3} i}$$ 
$$z^5=32\left(\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}}\right)$$ $$z=32\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$$

All in all, nicely done! Looks like you have a firm understanding of exponentation, polar and cartesian coordinates. The only thing I can point out is the minor algebra mistake right at the end. Notice that $\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(\frac{5\pi}{3}\right) =-\frac{\sqrt{3}}{2}.$ Hence for $z^5$ we have $$32\left(\cos\left(\frac{5\pi}{3}\right)+i\sin\left(\frac{5\pi}{3}\right)\right) = 32\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)  \\ = 16\left(1-i\sqrt{3}\right) $$
