Answer
$-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
