## College Algebra (10th Edition)

$-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)^2 ,$ use the square of a binomial and the equivalence $i^2=-1.$ $\bf{\text{Solution Details:}}$ 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 expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{1}{2}\right)^2+2\left( \dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}i \right)+\left(\dfrac{\sqrt{3}}{2}i \right)^2 \\\\= \dfrac{1}{4}+\dfrac{\sqrt{3}}{2}i+\dfrac{3}{4}i^2 .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{4}+\dfrac{\sqrt{3}}{2}i+\dfrac{3}{4}(-1) \\\\= \dfrac{1}{4}+\dfrac{\sqrt{3}}{2}i-\dfrac{3}{4} \\\\= -\dfrac{2}{4}+\dfrac{\sqrt{3}}{2}i \\\\= -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i .\end{array}