Adam had to find the trigonometric form of the complex number $\frac{1+3\mathrm{i}\sqrt3}{2-\mathrm{i}\sqrt3}$.
In which step of his solution did Adam make a mistake?
Adam's solution:
(1) Adam converted the complex number into its algebraic form. $$\frac{1+3\mathrm{i}\sqrt3}{2-\mathrm{i}\sqrt3}=\frac{(1+3\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}{(2-\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}=-1+\mathrm{i}\sqrt3$$
(2) Adam calculated the modulus of the complex number. $$|-1+\mathrm{i}\sqrt3|=\sqrt{(-1)^2+(\sqrt3)^2}=2$$
(3) Adam calculated the argument of the complex number. $$\sin\varphi=\frac{\sqrt3}{2}\implies\varphi=\frac{\pi}{3}$$
(4) Finally, Adam wrote the complex number in trigonometric form. $$\frac{1+3\mathrm{i}\sqrt3}{2-\mathrm{i}\sqrt3}=2\left(\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}\right)$$
In step (1). The correct simplification is
$$\frac{1+3\mathrm{i}\sqrt3}{2-\mathrm{i}\sqrt3}=\frac{(1+3\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}{(2-\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}=\frac{2+\mathrm{i}\sqrt3+6\mathrm{i}\sqrt3-9}{4-3}=-7+\mathrm{i}7\sqrt3.$$
In step (2). The correct calculation is $$|-1+\mathrm{i}\sqrt3|=\sqrt{-1^2+(\sqrt3)^2}=\sqrt2.$$
In step (3). The argument of the complex number must be a solution to the system of equations $$\sin\varphi=\frac{\sqrt3}{2}\wedge \cos\varphi=-\frac12.$$
The solutions are $\varphi=\frac{2\pi}{3}+2k\pi;\ k\in\mathbb{Z}$.
For example, the argument is $\varphi=\frac{2\pi}{3}$.
In step (4). The correct representation is $$2\left(\sin\frac{\pi}{3}+\mathrm{i}\cos\frac{\pi}{3}\right).$$
$$\frac{1+3\mathrm{i}\sqrt3}{2-\mathrm{i}\sqrt3}=\frac{(1+3\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}{(2-\mathrm{i}\sqrt3)(2+\mathrm{i}\sqrt3)}=-1+\mathrm{i}\sqrt3=2\left(\sin\frac{2\pi}{3}+\mathrm{i}\cos\frac{2\pi}{3}\right)$$