Answer
$\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$
Work Step by Step
Use special formula $(a-b)^2=a^2-2ab+b^2$.
We have $a= \frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}i $
$\left ( \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i \right )^2=\left ( \dfrac{\sqrt{3}}{2} \right )^2-2\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}i\right)+\left ( \dfrac{1}{2}i\right )^2$
Simplify.
$=\dfrac{3}{4}-\dfrac{\sqrt{3}}{2}i+\dfrac{1}{4}i^2 $
Use $i^2=-1$.
$=\dfrac{3}{4}-\dfrac{\sqrt{3}}{2}i-\dfrac{1}{4} $
Simplify.
$=\dfrac{3-1}{4}-\dfrac{\sqrt{3}}{2}i $
$=\dfrac{2}{4}-\dfrac{\sqrt{3}}{2}i $
$=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i $
Hence, the solution in the standard form is $\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$.
