Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form 
Convert complex to trig. 
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$

Let us consider 
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$ 
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{3}{3}\biggr) = \arctan (1)$$
$$\theta = \frac{\pi}{4}$$
Thus in trigonometric form we get
$$Z_1 = \biggr [3\sqrt2 \bigg(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\bigg)\bigg]^5$$
$$Z_1 =(3\sqrt 2)^5 \biggr [\bigg(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\bigg)\bigg]$$
Let us consider
$$(-2+2i)^3$$
$a = -2$, $b=2$
$$\sqrt{a^2+b^2}=\sqrt{(-2)^2+2^2} =2\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{2}{-2}\biggr) = \arctan (1)$$
$$\theta = \frac{3\pi}{4}$$
$$Z_2 =(2\sqrt 2)^3 \biggr [\bigg(\cos \frac{9\pi}{4}+i\sin \frac{9\pi}{4}\bigg)\bigg]$$
Multiplying $Z_1 \cdot Z_2$ we get
$$Z_1 \cdot Z_2 = (3\sqrt 2)^5 \cdot (2\sqrt2)^3  \biggr [\bigg(\cos \frac{5\pi}{4}+\cos \frac{9\pi}{4}\bigg)+i\sin \bigg(\frac{5\pi}{4}+i\sin\frac{9\pi}{4}\bigg)\bigg]$$ 
$$Z_1 \cdot Z_2 = 31104\biggr [\bigg(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\bigg)\bigg]$$
Let us consider
$$(\sqrt 3+i)^{10}$$
Here $a = \sqrt 3$, $b =1$
$$\sqrt{a^2+b^2}=\sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$$
$$\theta = \arctan \biggr(\frac{1}{\sqrt 3}\biggr) = \frac{\pi}{6}$$
$$Z_3 = 2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]$$
Now we have
$$\frac{Z_1 \cdot Z_2}{Z_3} = \frac{31104\biggr [\bigg(cos \biggr(\frac{7\pi}{2}\biggr) +i\sin \biggr(\frac{7\pi}{2}\biggr) \bigg]}{2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]} = \boxed {30.375 \biggr [ \cos \biggr(\frac{11\pi}{6}\biggr)+i\sin \biggr(\frac{11\pi}{6}\biggr)\biggr]}$$
Is my assumption correct?
 A: As an alternative by exponential form


*

*$(3+3i)^5=(3\sqrt 2e^{i \pi/4})^5$ 

*$(-2+2i)^3=(2\sqrt 2e^{i 3\pi/4} )^3$ 

*$(\sqrt 3 +i)^{10}=(2e^{i \pi/6} )^{10}$ 


then
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}=\frac{3^54\sqrt 2\cdot2^4\sqrt2}{2^{10}}\frac{e^{i 5\pi/4}\cdot e^{i 9\pi/4}}{e^{i 10\pi/6}}=30.375e^{i11\pi/6}$$
A: It might be useful to simplify before converting:
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}} = \frac{3^5\cdot 2^3\cdot ((1+i)(-1+i))^3(1+i)^2}{4^{10}}(\sqrt 3 - i)^{10}= \frac{3^5\cdot 2^3\cdot (-2)^3\cdot2i}{4^{10}}2^{10}e^{-i\frac{5}{3}\pi}$$$$= - \frac{243}{8}ie^{-i\frac{5}{3}\pi} =  -\frac{243}{8}e^{i\frac{5}{6}\pi}$$
A: I can't see any mistakes in your answers. Your calculations, however, are sometimes very wrong, not even leading the correct answers you pull out of them. 
For $z^3=(-2+2i)^3$, $r=2\sqrt{2}$ as you stated. $\theta=\tan^{-1}{(-1)}=\frac{-\pi}{4}$. The reason the correct result is $\frac{3\pi}{4}$, is due to $z$'s position on the argand diagram, the angle of $\frac{-\pi}{4}$ would create $z=2-2i$, adding $\pi$ rotates the point to $-2+2i$. 
When it comes to multiplying complex numbers, note this:
$$\cos A + \cos B \ne \cos (A+B)$$
Instead:
$$(\cos A+i\sin A)(\cos B+i\sin B)=\cos A \cos B-\sin A \sin B + i \sin A \cos B+ i\cos A \sin B)$$
$$=\cos (A+B)+i \sin (A+B)$$
Hence your overall result is correct, but your working is rather flawed.
