Consider three sets of complex roots of unity defined asA - zi - zi18 - 1B - zj - zj48 - 1 C - zizj - zi - A and zj - B- Then the number of distinct elements in C is

From De Moivre's Theorem, the nth root of unity is, z=cos(2kπn)+isin(2kπn) =cis(2kπn) Hence, 18th and 48th roots of unity are, cis(2π k_118) and cis(2π k_248) ...

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Standard XI

Mathematics

De-Moivre's Theorem

Consider thre...

Question

Consider three sets of complex roots of unity defined as
$A = {z_{i} : z_{i}^{18} = 1} B = {z_{j} : z_{j}^{48} = 1} C = {z_{i} z_{j} : z_{i} \in A and z_{j} \in B} .$
Then the number of distinct elements in $C$ is

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Solution

From De Moivre's Theorem, the $n th$ root of unity is,
$z = cos (\frac{2 k π}{n}) + i sin (\frac{2 k π}{n})$
$= cis (\frac{2 k π}{n})$

Hence, $18 th$ and $48 th$ roots of unity are,
$cis (\frac{2 π k_{1}}{18})$ and $cis (\frac{2 π k_{2}}{48})$ respectively,
where $k_{1}$ and $k_{2}$ are integers from $0$ to $17$ and $0$ to $47$ respectively.

$z_{i} z_{j} = cis (\frac{2 π k_{1}}{18} + \frac{2 π k_{2}}{48}) = cis (\frac{2 π (8 k_{1} + 3 k_{2})}{144})$
So, by De Moivre's Theorem, there are $144$ distinct elements in $C$ .

Alternate Solution:
$z_{i} z_{j} = cis (\frac{2 π k_{1}}{18} + \frac{2 π k_{2}}{48}) = cis (\frac{8 π k_{1} + 3 π k_{2}}{72})$
Since, $sin θ$ and $cos θ$ are periodic with period $2 π$ , there are at most $72 \times 2 = 144$ distinct elements in $C$ .

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Consider three sets of complex roots of unity defined as A={zi:z18i=1}B={zj:z48j=1}C={zizj:zi∈A and zj∈B}. Then the number of distinct elements in C is

Consider three sets of complex roots of unity defined as
$A = {z_{i} : z_{i}^{18} = 1} B = {z_{j} : z_{j}^{48} = 1} C = {z_{i} z_{j} : z_{i} \in A and z_{j} \in B} .$
Then the number of distinct elements in $C$ is