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

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(D) Given the equation $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.Taking the modulus on both sides,we get $|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$.Using

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$9$

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(D) Given the equation $(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$.Taking the modulus on both sides,we get $|(\sqrt{8} + i)^{50}| = |3^{49}(a + ib)|$.Using the property $|z^n| = |z|^n$,we have $|\sqrt{8} + i|^{50} = 3^{49} |a + ib|$.The modulus of $\sqrt{8} + i$ is $\sqrt{(\sqrt{8})^2 + 1^2} = \sqrt{8 + 1} = \sqrt{9} = 3$.Substituting this into the equation,we get $3^{50} = 3^{49} \sqrt{a^2 + b^2}$.Dividing both sides by $3^{49}$,we get $3 = \sqrt{a^2 + b^2}$.Squaring both sides,we obtain $a^2 + b^2 = 3^2 = 9$.

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

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