Could someone please help me write $z= \frac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form? Could someone please help me write $z = \dfrac{(2+2i)^3}{(1+i\sqrt{3})^4}$  on polar form?
$|z|=\sqrt{2}$
But how do I proceed to get the argument? 
$\mathrm{arg⁡}(z)=\mathrm{arg⁡}(2+2i)^3 - \mathrm{arg⁡}(1+i\sqrt{3})^4$
Thanks /David
 A: Hints: Use $\arctan 1=\frac{1}{4}\pi $ and $\arctan \sqrt{3}=\frac{1}{3}\pi $ in
$$\arg \left( \left( 2+2i\right) ^{3}\right) =3\arg \left( 2+2i\right)
=3 \arctan \frac{2}{2}=3\times \frac{1}{4}\pi $$
and
$$\arg \left( (1+i\sqrt{3})^{4}\right) =4\arg (1+i\sqrt{3})=4\arctan  \frac{\sqrt{3}}{1}=4\times \frac{1}{3}\pi .$$
A: Hint:  The argument of $1+i$ is $\frac{\pi}{4}$ by looking at the angle of the special triangle whose sides are $1$, $1$ and hypotenuse  $\sqrt{2}$. Also the argument of $1+i\sqrt{3}$ is $\frac{\pi}{3}$  by looking at the angle of the special triangle whose sides are $1$, $\sqrt{3}$ and hypotenuse $2$ .  Can you solve it from here?
A: You could just calculate and simplify the fraction:
$$ (2+2i)^3=8(1+i)^3=8\cdot (2i) \cdot (1+i)=-16+16i  $$
and
$$ (1+i\sqrt{3})^4=(-2+2\sqrt{3}i)^2=-8-8\sqrt{3}i$$
Amplify with the conjugate of the denominator and find the algebraic form of the complex number. Sometimes, when you don't remember the tricks presented in the other answers, it's easier to do the computation directly and only then transform into trigonometric form.
