If zk= cos ((k π/10))+i sin ((k π/10)), then z1 z2 z3 z4 is equal to

Question

If zk= cos ((k π/10))+i sin ((k π/10)), then z1 z2 z3 z4 is equal to (A) -1 (B) 2 (C) -2 (D) 1

Tardigrade Admin · Accepted Answer

We have zk=ωk where ω= cos ((π/10))+i sin ((π/10)) Thus, z1 z2 z3 z4=ω ⋅ ω2 ⋅ ω3 ⋅ ω4=ω10 = cos ((10 π/10))+i sin ((10 π/10)) [by the De Movire's Theorem] = cos π+i sin π=-1

Q. If $z_{k} = cos (\frac{kπ}{10}) + i sin (\frac{kπ}{10})$ , then $z_{1} z_{2} z_{3} z_{4}$ is equal to

-1

2

-2

1

We have $z_{k} = ω^{k}$ where
$ω = cos (\frac{π}{10}) + i sin (\frac{π}{10})$
Thus, $z_{1} z_{2} z_{3} z_{4} = ω \cdot ω^{2} \cdot ω^{3} \cdot ω^{4} = ω^{10}$
$= cos (\frac{10 π}{10}) + i sin (\frac{10 π}{10})$
[by the De Movire's Theorem]
$= cos π + i sin π = - 1$

Q. If zk​=cos(10kπ​)+isin(10kπ​), then z1​z2​z3​z4​ is equal to

-1

2

-2

1

We have zk​=ωk where ω=cos(10π​)+isin(10π​) Thus, z1​z2​z3​z4​=ω⋅ω2⋅ω3⋅ω4=ω10 =cos(1010π​)+isin(1010π​) [by the De Movire's Theorem] =cosπ+isinπ=−1

Q. If $z_{k} = cos (\frac{kπ}{10}) + i sin (\frac{kπ}{10})$ , then $z_{1} z_{2} z_{3} z_{4}$ is equal to

We have $z_{k} = ω^{k}$ where
$ω = cos (\frac{π}{10}) + i sin (\frac{π}{10})$
Thus, $z_{1} z_{2} z_{3} z_{4} = ω \cdot ω^{2} \cdot ω^{3} \cdot ω^{4} = ω^{10}$
$= cos (\frac{10 π}{10}) + i sin (\frac{10 π}{10})$
[by the De Movire's Theorem]
$= cos π + i sin π = - 1$