Question

The value of ${\left(1+i\right)}^5\times {\left(1-i\right)}^5$ is

• A. $-8$
• B. $8i$
• C. $8$
• D. $32$

Answer 1

Answer: (D)

$32$

We have,

${(1+i)}^5{(1-i)}^5$

$={(1+i)}^4{(1-i)}^4(1+i)(1-i)$

$={(1+i)}^4{(1-i)}^4(1^2-i^2)$

$={\left[{(1+i)}^2\right]}^2{\left[{(1-i)}^2\right]}^2(1^2-(-1))\therefore i^2=-1$

$={\left[(1^2+i^2+2i)\right]}^2{\left[(1^2+i^2-2i)\right]}^2\left(2\right)$

$=2{\left[(1^2+(-1)+2i)\right]}^2{\left[(1^2+(-1)-2i)\right]}^2\therefore i^2=-1$

$=2{\left[2i\right]}^2{[-2i]}^2$

$=32$

Hence, this is the answer.