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)

${\left(\frac{1+i}{1-i}\right)}^{1000}+{\left(\frac{1-i}{1+i}\right)}^{2000}$
$={\left(\frac{1+i}{1-i}\times \frac{1+i}{1+i}\right)}^{1000}+{\left(\frac{1+i^2-2i}{1-i^2}\right)}^{2000}$
$=i^{1000}+{\left(-i\right)}^{2000}$
$=i^{4\times 250}+{\left[{\left(-i\right)}^2\right]}^{4\times 250}$
$=1+1=2$