Question

If $z_r=\cos \frac{2r\pi }{5}+i\sin \frac{2r\pi }{5},r=0,1,3,4,....$, then $z_1{z}_2{z}_3{z}_4{z}_5$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

Converting in Eular's form we get

$z_1\cdot z_2\cdot z_3\cdot z_4\cdot z_5=1\cdot e^{i\frac{2\pi }{5}}\cdot e^{i\frac{4\pi }{5}}\cdot e^{i\frac{6\pi }{5}}\cdot e^{i\frac{8\pi }{5}}\cdot e^{i\frac{10\pi }{5}}$

$=1\cdot e^{i\frac{2\pi }{5}}\cdot e^{i\frac{4\pi }{5}}\cdot e^{i\frac{6\pi }{5}}\cdot e^{i\frac{8\pi }{5}}\cdot 1$

$=e^{i\frac{2\pi }{5}}\cdot e^{i\frac{4\pi }{5}}\cdot e^{i\frac{6\pi }{5}}\cdot e^{i\frac{8\pi }{5}}$

$=e^{i\frac{(2+4+6+8)\pi }{5}}$

$=e^{i4\pi }$

$=1$