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 $a=\frac{1}{2}$ and $b=\frac{\sqrt{3}}{2}i $
$\left ( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right )^2=\left ( \dfrac{1}{2}\right )^2+2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}i\right)+\left ( \dfrac{\sqrt{3}}{2}i \right )^2$
Simplify.
$=\dfrac{1}{4}+\dfrac{\sqrt{3}}{2}i+\dfrac{3}{4}i^2 $
Use $i^2=-1$.
$=\dfrac{1}{4}+\dfrac{\sqrt{3}}{2}i-\dfrac{3}{4} $
Simplify.
$=\dfrac{1-3}{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 $-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
