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Question

(A) Let $z = \sqrt{5} + 3i$. The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5} + 3i}$. To simplify,multiply the

Vedclass · Accepted Answer

$\frac{\sqrt{5}}{14}-\frac{3i}{14}$

Vedclass · Answer

(A) Let $z = \sqrt{5} + 3i$. The multiplicative inverse of $z$ is given by $z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5} + 3i}$. To simplify,multiply the numerator and denominator by the conjugate $\sqrt{5} - 3i$: $z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5} + 3i)(\sqrt{5} - 3i)}$. Using the identity $(a+bi)(a-bi) = a^2 + b^2$: $z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5})^2 + 3^2} = \frac{\sqrt{5} - 3i}{5 + 9} = \frac{\sqrt{5} - 3i}{14}$. Thus,$z^{-1} = \frac{\sqrt{5}}{14} - \frac{3i}{14}$.

Find the multiplicative inverse of the complex number $\sqrt{5}+3i$.

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