यदि x=frac{4}{5}+frac{3}{5} i और y=frac{sqrt{ | vedclass

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(A) दिया है $x = \frac{4+3i}{5}$,तो $\frac{1}{x} = \frac{5}{4+3i} = \frac{5(4-3i)}{16+9} = \frac{4-3i}{5}$.$x + \frac{1}{x} = \frac{4+3i+4-3i}{5} = \f

Vedclass · Accepted Answer

$\frac{-7 \sqrt{3}}{5 \sqrt{5}} i$

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(A) दिया है $x = \frac{4+3i}{5}$,तो $\frac{1}{x} = \frac{5}{4+3i} = \frac{5(4-3i)}{16+9} = \frac{4-3i}{5}$.$x + \frac{1}{x} = \frac{4+3i+4-3i}{5} = \frac{8}{5}$.$x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = (\frac{8}{5})^2 - 2 = \frac{64}{25} - 2 = \frac{14}{25}$.दिया है $y = \frac{\sqrt{3}-\sqrt{5}i}{\sqrt{8}}$,तो $\frac{1}{y} = \frac{\sqrt{8}}{\sqrt{3}-\sqrt{5}i} = \frac{\sqrt{8}(\sqrt{3}+\sqrt{5}i)}{3+5} = \frac{\sqrt{3}+\sqrt{5}i}{\sqrt{8}}$.$y + \frac{1}{y} = \frac{\sqrt{3}-\sqrt{5}i+\sqrt{3}+\sqrt{5}i}{\sqrt{8}} = \frac{2\sqrt{3}}{\sqrt{8}}$.$y - \frac{1}{y} = \frac{\sqrt{3}-\sqrt{5}i-(\sqrt{3}+\sqrt{5}i)}{\sqrt{8}} = \frac{-2\sqrt{5}i}{\sqrt{8}}$.$y^2 - \frac{1}{y^2} = (y + \frac{1}{y})(y - \frac{1}{y}) = (\frac{2\sqrt{3}}{\sqrt{8}})(\frac{-2\sqrt{5}i}{\sqrt{8}}) = \frac{-4\sqrt{15}i}{8} = \frac{-\sqrt{15}i}{2}$.अतः,$(x^2 + \frac{1}{x^2})(y^2 - \frac{1}{y^2}) = (\frac{14}{25})(\frac{-\sqrt{15}i}{2}) = \frac{-7\sqrt{15}i}{25} = \frac{-7\sqrt{3}\sqrt{5}i}{5\sqrt{5}\sqrt{5}} = \frac{-7\sqrt{3}}{5\sqrt{5}}i$.

यदि $x=\frac{4}{5}+\frac{3}{5} i$ और $y=\frac{\sqrt{3}}{\sqrt{8}}-\frac{\sqrt{5}}{\sqrt{8}} i$ है,तो $\left(x^2+\frac{1}{x^2}\right)\left(y^2-\frac{1}{y^2}\right)=$

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