Recall that
[tex]\cos\dfrac\pi6=\dfrac{\sqrt3}2[/tex]
[tex]\sin\dfrac\pi6=\dfrac12[/tex]
Then
[tex]2\sqrt3+2i=4\left(\dfrac{\sqrt3}2+\dfrac12i\right)=4\left(\cos\dfrac\pi6+i\sin\dfrac\pi6\right)=4e^{i\pi/6}[/tex]
