If (sqrt{8} + i)^{50} = 3^{49}(a + ib),then a | vedclass

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(C) Given: $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.Taking the modulus on both sides:$|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$$|\sqrt{8} + i|^{50} = 3^

Vedclass · Accepted Answer

$9$

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(C) Given: $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.Taking the modulus on both sides:$|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$$|\sqrt{8} + i|^{50} = 3^{49} |a + ib|$Since $|\sqrt{8} + i| = \sqrt{(\sqrt{8})^2 + 1^2} = \sqrt{8 + 1} = \sqrt{9} = 3$,we have:$3^{50} = 3^{49} \sqrt{a^2 + b^2}$Squaring both sides:$(3^{50})^2 = (3^{49})^2 (a^2 + b^2)$$3^{100} = 3^{98} (a^2 + b^2)$$a^2 + b^2 = \frac{3^{100}}{3^{98}} = 3^2 = 9$.

If $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$,then $a^2 + b^2$ is

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