#### Answer

$15-8i$

#### Work Step by Step

Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(4-i)^2
\\\\=
(4)^2-2(4)(i)+(i)^2
\\\\=
16-8i+i^2
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
16-8i+i^2
\\\\=
16-8i+(-1)
\\\\=
16-8i-1
\\\\=
15-8i
.\end{array}