Question

The value of $ \left(\frac{1+i}{1-i}\right)^{1000}+ \left(\frac{1-i}{1+i}\right)^{2000} $ is equal to

• A. $ 1+i $
• B. $ 2 $
• C. $ -i $
• D. $ 1 $

Answer 1

Answer: (B)

$\left(\frac{1+i}{1-i}\right)^{1000} + \left(\frac{ 1- i}{1+i}\right)^{2000}$
$= \left(\frac{ 1+i}{1-i} \times \frac{1+i}{1+i}\right)^{1000} + \left(\frac{1+i^{2} - 2i}{1-i^{2}}\right)^{2000} $
$ = i^{1000} + \left(-i\right)^{2000}$
$ = i^{4\times250} + \left[\left(-i\right)^{2}\right]^{4\times250} $
$= 1 + 1 = 2$