Let zk=cos((2 k π /10))+isin((2 k π /10)); textwherek=1,2, ldots 9 , then (|1 - z1 parallel 1 - z2| ldots |1 - z9|/10) equals

Question

Let zk=cos((2 k π /10))+isin((2 k π /10)); textwherek=1,2, ldots 9 , then (|1 - z1 parallel 1 - z2| ldots |1 - z9|/10) equals (A) 0 (B) 1 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

zk=ei (2 k π /10) z10-1=(z - 1)(z - z1)(z - z2) ldots (z - z9) undersetz arrow 1lim(z10 - 1/z - 1)= undersetz arrow 1lim(z - z1)(z - z2) ldots (z - z9) 10=(1 - z1)(1 - z2) ldots (1 - z9) 1=(|1 - z1| |1 - z2| ldots |1 - z9|/10)

Q. Let $z_{k} = cos (\frac{2 kπ}{10}) + i s in (\frac{2 kπ}{10}); where k = 1, 2, \dots 9$ , then $\frac{∣ 1 - z _{1} ∥ 1 - z _{2} ∣ \dots ∣ 1 - z _{9} ∣}{10}$ equals

$0$

$1$

$2$

$4$

Q. Let zk​=cos(102kπ​)+isin(102kπ​);wherek=1,2,…9 , then 10∣1−z1​∥1−z2​∣…∣1−z9​∣​ equals

0

1

2

4

zk​=ei102kπ​ z10−1=(z−1)(z−z1​)(z−z2​)…(z−z9​) z→1lim​z−1z10−1​=z→1lim​(z−z1​)(z−z2​)…(z−z9​) 10=(1−z1​)(1−z2​)…(1−z9​) 1=10∣1−z1​∣∣1−z2​∣…∣1−z9​∣​

Q. Let $z_{k} = cos (\frac{2 kπ}{10}) + i s in (\frac{2 kπ}{10}); where k = 1, 2, \dots 9$ , then $\frac{∣ 1 - z _{1} ∥ 1 - z _{2} ∣ \dots ∣ 1 - z _{9} ∣}{10}$ equals

$0$

$1$

$2$

$4$