Question

If ${{z}_{1}}$ and ${{z}_{2}}$ are the $nth$ roots of unity, then $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$ is a multiple of
A. $n\pi$
B. $\dfrac{3\pi }{n}$
C. $\dfrac{2\pi }{n}$
D. None of these.

Answer 1

Hint: To solve this question, we can represent ${{z}_{1}}={{e}^{i\dfrac{2k\pi }{n}}}$ and ${{z}_{2}}={{e}^{i\dfrac{2l\pi }{n}}}$ and then find $\dfrac{{{z}_{1}}}{{{z}_{2}}}$ in the form of ${{e}^{i\theta }}$. After that we can use the fact that $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\theta$ to get the desired result.

Complete step-by-step answer:

In this question, it is given that if ${{z}_{1}}$ and ${{z}_{2}}$ are the two $nth$ roots of unity, then we have to find $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$. Here ${{z}_{1}}$ and ${{z}_{2}}$ are in the form of complex numbers. We can write ${{z}_{1}}={{e}^{i\dfrac{2k\pi }{n}}}$ and ${{z}_{2}}={{e}^{i\dfrac{2l\pi }{n}}}$ as the general form of representing any complex number, that is the $nth$ roots of unity is given by, ${{e}^{i\dfrac{2r\pi }{n}}}$, where r is any integer. So, let us consider ${{z}_{1}}={{e}^{i\dfrac{2k\pi }{n}}}$ and ${{z}_{2}}={{e}^{i\dfrac{2l\pi }{n}}}$, where k and l are the integers. Now we have to find $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$.
We know that, ${{z}_{1}}={{e}^{i\dfrac{2k\pi }{n}}}$ and ${{z}_{2}}={{e}^{i\dfrac{2l\pi }{n}}}$, so by substituting it in the above, we get,
$\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{{{e}^{i\left( \dfrac{2k\pi }{n} \right)}}}{{{e}^{i\left( \dfrac{2l\pi }{n} \right)}}}\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}={{e}^{i\dfrac{2\left( k-l \right)\pi }{n}}}$, by using the fact that, $\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$.
Now, we will apply the fact that if $z={{e}^{i\theta }}$, then $\arg \left( z \right)=\theta$.
So, if $\dfrac{{{z}_{1}}}{{{z}_{2}}}={{e}^{i\dfrac{2\left( k-l \right)\pi }{n}}}$, then $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)=\dfrac{2\left( k-l \right)\pi }{n}$.
So, if k as well as l is an integer, then $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$ is represented as $\dfrac{2\pi }{n}\times$ any integer. So, we can say that the $\arg \left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)$ is a multiple of $\dfrac{2\pi }{n}$.
Hence, the correct answer is option C.

Note: An argument of a complex number, $z=x+iy$, denoted by $\arg \left( z \right)$ is defined in two equivalent ways:
(i) Geometrically, in a complex plane as the 2-D polar angle $\phi$ from a positive real axis to the vector representing t. The numeric angle is given by the angle in radius and is positive if measured anticlockwise.
(ii) Algebraically, as any real quantity $\phi$ such that $z=r\left( \cos \phi +\sin \phi \right)=r{{e}^{i\phi }}$ for some real r, where r is $\sqrt{{{x}^{2}}+{{y}^{2}}}$.