Aspire Faculty ID #15271 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

The amplitude of $\dfrac{1+i\sqrt{3}}{\sqrt{3}+i}$ is equal to …

Solution

$\dfrac{1+i\sqrt{3}}{\sqrt{3}+i} = \dfrac{(1+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} = \dfrac{(\sqrt{3}+3)i + (\sqrt{3}-1)}{4}$. Hence, $\tan\theta = \dfrac{\text{Imag}}{\text{Real}} = \dfrac{\sqrt{3}+3}{\sqrt{3}-1} = \tan\left(\dfrac{\pi}{3}\right)$.

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