Question

The center of a regular polygon of \[n\] sides is located at the point \[z=0\] and one of its vertices \[{{z}_{1}}\] is known. If \[{{z}_{2}}\] is the vertex adjacent to \[{{z}_{1}}\], then \[{{z}_{2}}\] is equal to
A. \[{{z}_{1}}(\cos \dfrac{2\pi }{n}\pm i\sin \dfrac{2\pi }{n})\]
B. \[{{z}_{1}}(\cos \dfrac{\pi }{n}\pm i\sin \dfrac{\pi }{n})\]
C. \[{{z}_{1}}(\cos \dfrac{\pi }{2n}\pm i\sin \dfrac{\pi }{2n})\]
D. None of these

Answer 1

Hint: In this question, we have to find the unknown vertex in the given regular polygon. Since the vertices are complex, Euler’s formula and the mod amplitude form of a complex number is used for finding 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, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
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: Given that,
One of the vertices of a regular polygon is known. I.e., \[{{z}_{1}}\]
The required vertex is adjacent to the known vertex. I.e., \[{{z}_{2}}\]
There are two possibilities of \[{{z}_{2}}\] i.e., \[{{z}_{2}}\] can be obtained by rotating \[{{z}_{1}}\] through $\dfrac{2\pi }{n}$ either in clockwise or in anticlockwise direction.
So, we can write
$\dfrac{{{z}_{2}}}{{{z}_{1}}}=\left| \dfrac{{{z}_{2}}}{{{z}_{1}}} \right|{{e}^{\pm i\theta }}$
where $\theta =\dfrac{2\pi }{n}$ and \[\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|\]
Then,
$\begin{align}
  & \dfrac{{{z}_{2}}}{{{z}_{1}}}={{e}^{\pm i\dfrac{2\pi }{n}}} \\
 & \Rightarrow {{z}_{2}}={{z}_{1}}{{e}^{\pm i\dfrac{2\pi }{n}}} \\
 & \Rightarrow {{z}_{2}}={{z}_{1}}\left( \cos \dfrac{2\pi }{n}\pm i\sin \dfrac{2\pi }{n} \right) \\
\end{align}$\[\]

Option ‘A’ is correct
Note: Here we need to remember that, the required vertex \[{{z}_{2}}\] is possible to be adjacent in two ways i.e., either in a clockwise direction or in the anticlockwise direction of \[{{z}_{1}}\].