The magnitude of $I$ is given by $\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$. The angle $\theta$ for $I$ is given by $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^{\circ}$. Therefore, $I$ in trigonometric form is $2 \operatorname{cis} 30^{\circ}$.
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