Polar form of $\left(2\sqrt3-2i\right)\left(-2\sqrt3+6i\right)$ is-

$a+ib=r(\cos{\theta}+i\sin{\theta})$ then $a+ib=re^{i\theta}$ [euler’s formula]

modulus of $2\sqrt3-2i, r_1=\sqrt{\left(2\sqrt3\right)^2+\left(-2\right)^2}=4$

argument, $\theta_1=\tan^{-1}{\left(-\frac{2}{2\sqrt3}\right)}=-\frac{\pi}{6}\ \ \therefore2\sqrt3-2i=4e^{-i\frac{\pi}{6}}\ldots\ldots\ldots\left(i\right)$

modulus of $-2\sqrt3+6i, r_2=\sqrt{\left(-2\sqrt3\right)^2+\left(6\right)^2}=4\sqrt3$

argument, $\theta_2=\pi-\tan^{-1}{\left|\frac{6}{-2\sqrt3}\right|}=\pi-\tan^{-1}{\sqrt3}=\frac{2\pi}{3}\ \ \therefore-2\sqrt3+6i=4\sqrt3e^{i\frac{2\pi}{3}}$

$\therefore\left(2\sqrt3-2i\right)\left(-2\sqrt3+6i\right)=4e^{-i\frac{\pi}{6}}\times4\sqrt3e^{i\frac{2\pi}{3}}=16\sqrt3e^{\left(-\frac{\pi}{6}+\frac{2\pi}{3}\right)i}=16\sqrt3e^{i\frac{\pi}{2}}$

Alternate: Use calculator.