Find multiplicative inverse of the complex number $(2 + \sqrt{3 i})^{2} .$

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Solution

Let z= $(2 + \sqrt{3 i)^{2}},$ then $z = 4 + 3 i^{2} + \sqrt{3 i} = 4 - 3 + \sqrt{3 i} = 1 + 4 \sqrt{3 i}$

$∴$ Multiplicative inverse of $z = \frac{1}{z} = \frac{1}{1 + 4 \sqrt{3 i}} = \frac{1 - 4 \sqrt{3 i}}{(1 + 4 \sqrt{3 i}) (1 - 4 \sqrt{3 i})} = \frac{1 - 4 \sqrt{3 i}}{1 + 48}$

= $\frac{1}{49} - \frac{4 \sqrt{3 i}}{49}$

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