{left( frac{1 + i}{1 - i} right)^2} + {left( | 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

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) First,simplify the term $\frac{1 + i}{1 - i}$ by multiplying the numerator and denominator by the conjugate $(1 + i)$:$\frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$.Similarly,$\frac{1 - i}{1 + i} = \frac{1}{i} = -i$.Now,substitute these values into the expression:$(i)^2 + (-i)^2 = -1 + (-1)^2 = -1 + 1 = 0$.Wait,re-evaluating the expression:${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2} = (i)^2 + (-i)^2 = -1 + (-1) = -2$.

${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2}$ is equal to

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