left(frac{1+i}{1-i}right)^{228} = | vedclass

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(C) First,simplify the expression inside the parenthesis: $\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+

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$\left(\frac{1-i}{1+i}\right)^{228}$

Vedclass · Answer

(C) First,simplify the expression inside the parenthesis: $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$. Now,substitute this into the original expression: $i^{228} = (i^4)^{57} = 1^{57} = 1$. Next,evaluate the options to see which one equals $1$: For option $C$: $\left(\frac{1-i}{1+i}\right)^{228} = \left(\frac{1}{i}\right)^{228} = \frac{1}{i^{228}} = \frac{1}{1} = 1$. Thus,the correct option is $C$.

$\left(\frac{1+i}{1-i}\right)^{228} = $

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