Let ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ be their vertices of an equilateral triangle circumscribing the circle $\left| z \right|=\dfrac{1}{2}$. If ${{z}_{1}}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$ and ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ are in an anticlockwise sense then ${{z}_{2}}$ is
a. $1+\sqrt{3}i$
B. $1-\sqrt{3}i$
C. $1$
D. $-1$

Hint: In this question, we have to find the vertex ${{z}_{2}}$ of the given equilateral triangle. Since the given triangle is an equilateral triangle, the three angles are equal and $\theta =\dfrac{\pi }{3}$. By using this, we can able to find the required vertex.

Formula Used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$

Complete step by step solution: The given equilateral triangle’s three vertices are ${{z}_{1}},{{z}_{2}},{{z}_{3}}$
The vertex ${{z}_{1}}$ is given as ${{z}_{1}}=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
Since the vertex ${{z}_{2}}$ is adjacent to the vertex ${{z}_{1}}$, we can write
${{z}_{2}}={{z}_{1}}{{e}^{i2\theta }}$
Where the angle $\theta =\dfrac{\pi }{3}$
So,
$\begin{align}
  & {{z}_{2}}={{z}_{1}}{{e}^{i2\theta }} \\
 & \text{ }={{z}_{1}}{{e}^{i\dfrac{2\pi }{3}}} \\
 & \text{ }=\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right){{e}^{i\dfrac{2\pi }{3}}} \\
\end{align}$
On expanding the Euler form, we get
$\begin{align}
  & {{z}_{2}}=\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)\left( \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right) \\
 & \text{ }=\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)\left( \dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}i \right) \\
 & \text{ }=\dfrac{-1}{4}-\dfrac{3}{4} \\
 & \text{ }=-1 \\
\end{align}$

Option ‘D’ is correct

Note: Here we need to remember that the given triangle is an equilateral triangle. So, all three angles are equal and are given by $\theta =\dfrac{\pi }{3}$. By using this, we can find the required vertex.