Question

Let $z_{k}=cos\left(\frac{2 k \pi }{10}\right)+isin\left(\frac{2 k \pi }{10}\right);\text{where}k=1,2,\ldots 9$ , then $\frac{\left|1 - z_{1} \parallel 1 - z_{2}\right| \ldots \left|1 - z_{9}\right|}{10}$ equals

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$z_{k}=e^{i \frac{2 k \pi }{10}}$
$z^{10}-1=\left(z - 1\right)\left(z - z_{1}\right)\left(z - z_{2}\right)\ldots \left(z - z_{9}\right)$
$\underset{z \rightarrow 1}{lim}\frac{z^{10} - 1}{z - 1}=\underset{z \rightarrow 1}{lim}\left(z - z_{1}\right)\left(z - z_{2}\right)\ldots \left(z - z_{9}\right)$
$10=\left(1 - z_{1}\right)\left(1 - z_{2}\right)\ldots \left(1 - z_{9}\right)$
$1=\frac{\left|1 - z_{1}\right| \left|1 - z_{2}\right| \ldots \left|1 - z_{9}\right|}{10}$