If z 1- z 2- z 3- n n are nth- roots of unity- then for k -1-2- - - - nA- - z k -1- k - z k -B- - z k -1- z k - z k -1-C- - z k - z k -1-D- - z k - k - z k -1-

The n^th roots of unity are given by z_k = e^i2π(k 1)/n, (k = 1, 2,....., n) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quadratic Polynomial

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, z_{3} . . . . . n_{n}$ are n^th, roots of unity, then for k = 1, 2, ....., n

| z_{k} | = k | z_{k + 1} |

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| z_{k + 1} | = k | z_{k} |

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| z_{k + 1} | = | z_{k} | + | z_{k + 1} |

No worries! We‘ve got your back. Try BYJU‘S free classes today!

| z_{k} | = | z_{k + 1} |

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

z_{k} = e^{i θ k}

k = 1, 2, 3, 4

i^{2} = - 1

∣ ∣ ∣ 4 \sum k = 1 \frac{1}{z_{k}} ∣ ∣ ∣ = 1,

∣ ∣ ∣ 4 \sum k = 1 z_{k} ∣ ∣ ∣

z_{k} = c o s (\frac{2 k π}{10}) + i s i n (\frac{2 k π}{10}); k = 1, 2, . . . . . . . ., 9.

	List - I		List - II
(P)	For each $z_{k}$ there exists a $z_{j}$ such $z_{k} . z_{j} =$ 1	(1)	True
(Q)	There exists a k $\in$ {1, 2, ........, 9} such that $z_{1} \cdot z = z_{k}$ has no solution z in the set of complex numbers	(2)	False
(R)	$\frac{\| 1 - z_{1} \| 1 - z_{2} \| . . . . . \| 1 - z_{9} \|}{10}$ equal	(3)	1
(S)	$1 - \sum_{k = 1}^{9} c o s (\frac{2 k π}{10})$	(4)	2