Answer
$-7+4i\sqrt{11}$
Work Step by Step
RECALL:
(1) $(a-b)^2 = a^2-2ab+b^2$
(2) $\sqrt{-1}=i$
(3) $i^2=-1$
(4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$.
Use rule (4) above to obtain:
$=(-2-i\sqrt{11})^2$
Use rule (1) above with $a=-2$ and $b=i\sqrt{11}$ to obtain:
$=(-2)^2-2(-2)(i\sqrt{11}) + (i\sqrt{11})^2
\\=4+4i\sqrt{11}+i^2(11)
\\=4+4i\sqrt{11} + 11i^2$
Use rule (3) above to obtain:
$=4+4i\sqrt{11} + 11(-1)
\\=4+4i\sqrt{11} -11
\\=-7+4i\sqrt{11}$
