The exercise is:

write the expression in polar form: $\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$

Here is my work:

$\displaystyle \frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$

that is

$\displaystyle \frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2 })(2\sqrt{3}-2i)}$

and that becomes:

$\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$

and then

$\displaystyle \frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$

$\displaystyle \frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{ 3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}$

anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.?