(1+i)^{2024}+(1-i)^{2024} = | vedclass

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(B) We know that $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$ and $(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$. Substituting these into the expression: $(1+i)^{2024} + (

Vedclass · Accepted Answer

$2^{1013}$

Vedclass · Answer

(B) We know that $(1+i)^2 = 1+i^2+2i = 1-1+2i = 2i$ and $(1-i)^2 = 1+i^2-2i = 1-1-2i = -2i$. Substituting these into the expression: $(1+i)^{2024} + (1-i)^{2024} = [(1+i)^2]^{1012} + [(1-i)^2]^{1012}$ $= (2i)^{1012} + (-2i)^{1012}$ $= 2^{1012} \cdot i^{1012} + (-2)^{1012} \cdot i^{1012}$ Since $1012$ is an even number,$(-2)^{1012} = 2^{1012}$ and $i^{1012} = (i^4)^{253} = 1^{253} = 1$. $= 2^{1012} \cdot 1 + 2^{1012} \cdot 1$ $= 2 \cdot 2^{1012} = 2^{1013}$.

$(1+i)^{2024}+(1-i)^{2024} = $

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