(C) We know that $(1 + i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i$. Similarly,$(1 - i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$. Now,$(1 + i)^8 + (1 - i)^8 = (

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(C) We know that $(1 + i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i$. Similarly,$(1 - i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$. Now,$(1 + i)^8 + (1 - i)^8 = ((1 + i)^2)^4 + ((1 - i)^2)^4$. Substituting the values,we get $(2i)^4 + (-2i)^4$. This simplifies to $16i^4 + 16i^4$. Since $i^4 = 1$,we have $16(1) + 16(1) = 16 + 16 = 32$.