Question

If p and q are distinct prime numbers, then find the number of distinct imaginary numbers which are $p^(th)$ as well $q^(th)$ roots of unity.
A) min (p,q)
B) max (p,q)
C) 1
D) zero

Answer 1

Hint: In \[C\] (\[C\] is a set of complex numbers ), given z $ \ne $ 0, the equation expansion ${w^n} = z$has n- distinct solution given by ${w_k} = \sqrt[n]{r}.{e^{i\left( {\dfrac{{\theta + 2k\pi }}{n}} \right)}}$ , k = 0,1,2,………, n-1 where r=$\left| z \right|$ and $\theta $= arg z
Also, ${x^n} = 1$ means nth root of unity.

Complete step by step answer:
Now p and q are distinct prime numbers means they are not same and their HCF is 1
So, using hint we can say that pth root of unity means ${x^p} = 1$
$\therefore {x^p} - 1 = 0$
Also, we can say that qth root of unity means \[{x^q} = 1\]
 $\therefore {x^q} - 1 = 0$
Now the roots of p using the hint are 1, ${e^{2\pi i}}$,${e^{4\pi i}}$………. ,(P-1)
Therefore, $x = {(1)^{\dfrac{1}{p}}},{e^{\dfrac{{2\pi i}}{p}}},{e^{\dfrac{{4\pi i}}{p}}},{e^{\dfrac{{8\pi i}}{p}}},........,{e^{\dfrac{{(p - 1)\pi i}}{p}}}$
Now the roots of q using the hint are 1, ${e^{2\pi i}}$,${e^{4\pi i}}$………. ,(q-1)
Therefore, $x = {(1)^{\dfrac{1}{q}}},{e^{\dfrac{{2\pi i}}{q}}},{e^{\dfrac{{4\pi i}}{q}}},{e^{\dfrac{{8\pi i}}{q}}},........,{e^{\dfrac{{(q - 1)\pi i}}{q}}}$
Also, p and q are prime numbers so their factors are 1 & p and 1 & q respectively
Therefore, all powers in the term of i are going to be unique.
Therefore, the only common root is 1 between them and which implies they do not have any common root which are imaginary numbers.

Hence, option D is correct.

Note:
We can make use of De Moivre’s theorem to find nth roots of unity in C
De Moivre’s theorem: If n is any integer or fraction then ${\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta $