Given the following complex numbers: $$ z=1+i\sqrt{3} \qquad w = 0.707 - 0.707i $$ find the cartesian forms of the following expressions: $$ z^2 \bar{w}\qquad\text{and}\qquad \frac{z^3}{w^9} $$
The first one i found the answer to be 1.414 - 1.414i, is this correct?
Hint:
Observe that $\;0.707\cong\cfrac1{\sqrt2}\approx\cfrac{\sqrt2}2\;$ , thus
$$w\approx\frac{\sqrt2}2(1-i)$$
Thus the first one is
$$z^2\overline w=(-2+2\sqrt3\,i)\frac1{\sqrt2}(1+i)=\sqrt2(-1+\sqrt3~i)(1+i)=\sqrt2(-1-\sqrt3-(1-\sqrt3)i)=$$
$$=-\sqrt2\left[(1+\sqrt3+(1-\sqrt3)~\right]$$
Continue from here...
As commented, in polar form is way easier. Choosing the standard argument in [0,2\pi);$ , we get:
$$\;z=2e^{\pi i/4}\;,\;\;w=e^{3\pi i/4}\implies\text{etc.}$$
and etc.
If you know the polar forms of $z$ and $w$, then the calculation becomes much easier as mentioned in the comments.
Suppose that $z=r_{1}e^{i\theta_{1}}$ and $w=r_{2}e^{i\theta_{2}}$, then
$z^{2}\bar{w}=(r_{1}e^{i\theta_{1}})^{2}(r_{2}e^{-i\theta_{2}})=r_{1}^{2}r_{2}e^{i(2\theta_{1}-\theta_{2})}$ and $\frac{z^{3}}{w^9}=\frac{r_{1}^{3}}{r_{2}^{9}}e^{3i(\theta_{1}-3\theta_{2})}.$
Then you can use Euler's formula: $re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$ to express your result in the form $x+iy.$
Now since $r_{1}=|z|=2$ and $\theta_{1}=\arctan(\frac{\sqrt{3}}{1})=\frac{\pi}{3}$ you have that $z=2e^{i\frac{\pi}{3}}.$ Similarly $w=r_{2}e^{-i\frac{\pi}{4}}$ where $r_{2}=|w|=\sqrt{(0.707)^2+(0.707)^2}\space(\approx 0.9998489885978).$
