Let zk - cos2k-10 -i sin2k-10 - k- 1- 2- - 9- List I List II- P- For each zk there exist a zj 1 True- such that zk - zj - 1- Q- There exists a k -1- 2- - 9- such that z1 - z - z k has no solution 2 False- z in the set of complex numbers- R- -1-z1-1-z2-1 -z9-10 equals 3 1- S- 1 - -k - 19cos2k-10 equals 4 2- -Which of the following option is correct-

We know that, z _k nbsp; = e ^ik(2π/10) (P) Assuming z_j = e ^ij(2π/10) Now, z_k · z_j = 1 ⇒ nbsp;e ^i(k+j)2π/10= 1 ⇒cos(2(k+j)π10)=1 ⇒ nbsp;2(k+j)π10=2n ...

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

Byju's Answer

Standard XII

Mathematics

Properties of Argument

Let zk = cos2...

Question

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9.$

$\begin{matrix} L i s t (I) & L i s t (I I) P . & For each z_{k} there exist a z_{j} & (1) & True such that z_{k} \cdot z_{j} = 1 Q . & There exists a k \in {1, 2, \dots, 9} such that z_{1} \cdot z = z_{k} has no solution & (2) & False z in the set of complex numbers. R . & \frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10} equals & (3) & 1 S . & 1 - 9 \sum k = 1 cos (\frac{2 k π}{10}) equals & (4) & 2 \end{matrix}$

Which of the following option is correct?

(P) \to (1), (Q) \to (2) (R) \to (4), (S) \to (3)

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

(P) \to (2), (Q) \to (1) (R) \to (3), (S) \to (4)

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

(P) \to (1), (Q) \to (2) (R) \to (3), (S) \to (4)

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

(P) \to (2), (Q) \to (1) (R) \to (4), (S) \to (3)

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

Open in App

Solution

The correct option is C $(P) \to (1), (Q) \to (2) (R) \to (3), (S) \to (4)$
We know that,
$z_{k} = e^{i k (2 π / 10)}$

$(P)$
Assuming
$z_{j} = e^{i j (2 π / 10)}$
Now,
$z_{k} \cdot z_{j} = 1 \Rightarrow e^{i (k + j) 2 π / 10} = 1 \Rightarrow cos (\frac{2 (k + j) π}{10}) = 1 \Rightarrow \frac{2 (k + j) π}{10} = 2 n π, n \in Z$
When $n = 1$
$\Rightarrow \frac{2 π (j + k)}{10} = 2 π \Rightarrow j = 10 - k$
Hence, True

$(Q)$
$z_{1} \cdot z = z_{k} \Rightarrow z = \frac{z_{k}}{z_{1}} = e^{i (k - 1) 2 π / 10}$
There exist such $z$ for different values of $k$
Hence, false

$(R)$
We know that,
if $z_{k}$ are roots of $z^{10} - 1 = 0$
$z^{10} - 1 = (z - 1) (z - z_{1}) (z - z_{2}) \dots (z - z_{9}) \Rightarrow \frac{z^{10} - 1}{z - 1} = (z - z_{1}) (z - z_{2}) \dots (z - z_{9}) \Rightarrow z^{9} + z^{8} + \dots + z^{1} + 1 = (z - z_{1}) (z - z_{2}) \dots (z - z_{9})$
Putting $z = 1$
$(z - z_{1}) (z - z_{2}) \dots (z - z_{9}) = 10 \Rightarrow \frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10} = 1$

$(S)$
$z^{10} - 1 = 0$
$\Rightarrow (z - 1) (z - z_{1}) (z - z_{2}) \dots (z - z_{9}) = 0$
Hence sum of the roots
$1 + z_{1} + z_{2} + \dots + z_{8} + z_{9} = 0$
$\Rightarrow 1 + 9 \sum k = 1 cos (\frac{2 k π}{10}) + i 9 \sum k = 1 sin (\frac{2 k π}{10}) = 0 \Rightarrow 9 \sum k = 1 cos (\frac{2 k π}{10}) = - 1; 9 \sum k = 1 sin (\frac{2 k π}{10}) = 0$
Therefore,
$1 - 9 \sum k = 1 cos (\frac{2 k π}{10}) = 1 + 1 = 2$

Hence,
$(P) \to (1), (Q) \to (2) (R) \to (3), (S) \to (4)$

Suggest Corrections

Similar questions

Q. Let

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