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If ‘p’ and ‘q’ are distinct prime numbers, than the number of distinct imaginary numbers which are P^{th}  as well as q^{th} roots of unity are -

Option: 1

min ^m ()p,q)


Option: 2

max^m ()p,q)


Option: 3


Option: 4


Answers (1)

best_answer

As we have learnt in

 

nth roots of unity -

z=\left ( 1 \right )^{\frac{1}{n}}\Rightarrow z=\cos \frac{2k\pi }{n}+i\sin \frac{2k\pi }{n}

Where k=0,1,2,......,(n-1)

-

 

 

\large x^{p}-1=0 \qquad {\color{Red} x^{q}-1=0}

\large x^{p}=1, e^{2 \pi i}, e^{4 \pi i}\ldots\ldots p^{th} \qquad {\color{Red} x^{q}=1, e^{2 \pi i}, e^{4 \pi i}\ldots\ldots q^{th}}

\large x=(1)^{\frac1p}, e^{ \frac{2 \pi i}{p}}, e^{ \frac{4\pi i}{p}} \cdots p^{th}\qquad {\color{Red} x=(1)^{\frac1q}, e^{\frac{2 \pi i}{q}}, e^{ \frac{4\pi i}{q}} \cdots q^{th}}

p and q are prime numbers

Hence their factors will be 

p = 1 X p

q = 1 X q

Therefore

\large e^{ \frac{2 \pi i}{p}}\neq {\color{Red} e^{\frac{2 \pi i}{q}}}

\large e^{ \frac{4 \pi i}{p}}\neq {\color{Red} e^{\frac{4 \pi i}{q}}}

.

.

.

and so on..

Since

\large 1= {\color{Red} 1}

and it is not an imaginary root.

Posted by

shivangi.shekhar

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