$|z| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2} =$ $\sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\theta = \arctan\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} =$ $\arctan\frac{\sqrt{3}}{2}2 = \arctan\sqrt{3} = \frac{\pi}{3}$