left(frac{1+i}{1-i}right)^4+left(frac{1-i}{1+ | vedclass

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Question

(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $1+i$: $\frac{1+i}{1-i} = \frac{(1+i)(1+i)

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$2$

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(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $1+i$: $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1+1} = \frac{2i}{2} = i$. Similarly,for the second term: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1+i^2-2i}{1-i^2} = \frac{1-1-2i}{1+1} = \frac{-2i}{2} = -i$. Now,substitute these values into the original expression: $(i)^4 + (-i)^4 = i^4 + i^4 = 1 + 1 = 2$.

$\left(\frac{1+i}{1-i}\right)^4+\left(\frac{1-i}{1+i}\right)^4$ is equal to

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