Answer
$12-16i$
Work Step by Step
We are given the expression $(4-2i)^{2}$. Found on page 285, we know that the square of a binomial $(a-b)^{2}$ simplifies to $a^{2}-2ab+b^{2}$.
Therefore, $(4-2i)^{2}=(4)^{2}-2(4)(2i)+(2i)^{2}=16-16i+4i^{2}=16-16i+(4\times-1)=16-16i-4=12-16i$
We know that $i^{2}=-1$.