Question

If $n$ is an odd integer, then $(1+i{)}^{6n}+(1-i{)}^{6n}=$

• A. $2i$
• B. $-2i$
• C. $0$
• D. $-2$

Answer 1

The correct option is C $0$

$(1+i{)}^{6n}+(1-i{)}^{6n}$
$={\left[\right(1+i{)}^2]}^{3n}+{\left[\right(1-i{)}^2]}^{3n}$
$=(2i{)}^{3n}+(-2i{)}^{3n}$
$=2^{3n}[i^{3n}+(-i{)}^{3n}]$
$=2^{3n}[i^{3n}-i^{3n}]$ as $n$ is odd.
$=0$