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(A) Given expression: $\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$Using the identity $(a+b)(a-b) = a^2-b^2$ in the n

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$\frac{-7\sqrt{2}i}{2}$

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(A) Given expression: $\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$Using the identity $(a+b)(a-b) = a^2-b^2$ in the numerator:$= \frac{3^2 - (i\sqrt{5})^2}{\sqrt{3} + \sqrt{2}i - \sqrt{3} + i\sqrt{2}}$$= \frac{9 - 5i^2}{2\sqrt{2}i}$Since $i^2 = -1$:$= \frac{9 - 5(-1)}{2\sqrt{2}i} = \frac{14}{2\sqrt{2}i} = \frac{7}{\sqrt{2}i}$Multiply numerator and denominator by $i$:$= \frac{7i}{\sqrt{2}i^2} = \frac{7i}{\sqrt{2}(-1)} = \frac{-7i}{\sqrt{2}}$Rationalizing the denominator:$= \frac{-7i 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}} = \frac{-7\sqrt{2}i}{2}$

Express the following expression in the form of $a+ib$:
$\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

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Express the following expression in the form of $a+ib$: $\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$