How to calculate it? :$|V_p|(1∠0^\circ-1∠-120^\circ)=\sqrt{3}|V_p|∠30^\circ$

Question

I saw this math formula in the circuit system. It said $$|V_p|(1∠0^\circ-1∠-120^\circ)=\sqrt{3}|V_p|∠30^\circ$$

Does anyone know how to calculate $|V_p|(1∠0^\circ-1∠-120^\circ)$ to $\sqrt{3}|V_p|∠30^\circ$ ?

Example of the notation: $$6∠60^\circ=(6 \cos 60^\circ)+j(6 \sin 60^\circ)$$

Answer 1

Simply add them $$1\angle0°=1\cos0°+j1\sin0°=1+j0$$ $$1\angle-120°=1\cos-120°+j1\sin-120°=-\frac12-j\frac{\sqrt3}2$$ $$1\angle0-1\angle-120°=\frac32+j\frac{\sqrt3}2$$ If we isolated the $\sqrt3$, we have $$\sqrt3\left(\frac{\sqrt3}2+j\frac12\right)=\sqrt3\left(\cos30°+j\sin30°\right)=\sqrt3\angle30°$$