frac{(1+i)^{2016}}{(1-i)^{2014}} is equal to | vedclass

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(A) We have,$\frac{(1+i)^{2016}}{(1-i)^{2014}} = \frac{(1+i)^{2014} 	imes (1+i)^2}{(1-i)^{2014}}$ $= \left(\frac{1+i}{1-i}ight)^{2014} 	imes (1+i)

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$-2i$

Vedclass · Answer

(A) We have,$\frac{(1+i)^{2016}}{(1-i)^{2014}} = \frac{(1+i)^{2014} 	imes (1+i)^2}{(1-i)^{2014}}$ $= \left(\frac{1+i}{1-i}ight)^{2014} 	imes (1+i)^2$ Since $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{2i}{2} = i$ So,the expression becomes $i^{2014} 	imes (1+2i+i^2)$ $= i^{2014} 	imes (2i)$ Since $i^4 = 1$,$i^{2014} = (i^4)^{503} 	imes i^2 = 1^{503} 	imes (-1) = -1$ Therefore,$-1 	imes 2i = -2i$

$\frac{(1+i)^{2016}}{(1-i)^{2014}}$ is equal to

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