(C) Let $z = \sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1}$.First,calculate $z^2 = (\sqrt{2}+1) - (\sqrt{2}-1) + 2i\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)}$.$z^2 = 2

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(C) Let $z = \sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1}$.First,calculate $z^2 = (\sqrt{2}+1) - (\sqrt{2}-1) + 2i\sqrt{(\sqrt{2}+1)(\sqrt{2}-1)}$.$z^2 = 2 + 2i\sqrt{2-1} = 2 + 2i$.Now,$z^8 = (z^2)^4 = (2 + 2i)^4$.$z^8 = [2(1+i)]^4 = 16(1+i)^4$.Since $(1+i)^2 = 1 + 2i + i^2 = 2i$,we have $(1+i)^4 = (2i)^2 = 4i^2 = -4$.Therefore,$z^8 = 16 \times (-4) = -64$.

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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$(\sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1})^8 =$

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A
$64$
B
$64i$
C
$-64$
D
$-64i$

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