Question

If $z_r=\cos \frac{2\pi r}{5}+i\sin \frac{2\pi r}{5},r=0,1,2,3,4$ then $z_0{z}_1{z}_2{z}_3{z}_4$ is equal to

• A. - 1
• B. 0
• C. 1
• D. none of these

Answer 1

Answer: (C)

1

GIven that,

$z_r=\cos \left(\frac{2\pi r}{5}\right)+i\sin \left(\frac{2\pi r}{5}\right)$

To find $z_0{z}_1{z}_2{z}_3{z}_4$

$z_0=\cos 0+i\sin 0=1$

$z_1=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}$

$z_2=\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}$

$=\cos (\pi -\frac{\pi }{5})+i\sin (\pi -\frac{\pi }{5})$

$z_2=-\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}$

$z_3=\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}$

$=\cos (\pi +\frac{\pi }{5})+i\sin (\pi +\frac{\pi }{5})$

$z_3=-\cos \frac{\pi }{5}-i\sin \frac{\pi }{5}$

$z_4=\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}=\cos (2\pi -\frac{2\pi }{5})+i\sin (2\pi -\frac{2\pi }{5})$

$z_4=\cos \frac{2\pi }{5}-i\sin \frac{2\pi }{5}$

Now $z_0{z}_1{z}_2{z}_3{z}_4$$=1(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5})(-\cos \frac{\pi }{5}+i\sin \frac{\pi }{5})$

$(-\cos \frac{\pi }{5}-i\sin \frac{\pi }{5})(\cos \frac{2\pi }{5}-i\sin \frac{2\pi }{5})$

$({\cos }^2\frac{2\pi }{5}-i^2{\sin }^2\frac{2\pi }{5})[-(\cos \frac{\pi }{5}-i\sin \frac{\pi }{5})\times -(\cos \frac{\pi }{5}+i\sin \frac{\pi }{5})]$

$=({\cos }^2\frac{2\pi }{5}+{\sin }^2\frac{2\pi }{5})({\cos }^2\frac{\pi }{5}-i^2{\sin }^2\frac{\pi }{5})$

$=({\cos }^2\frac{2\pi }{5}+{\sin }^2\frac{2\pi }{5})({\cos }^2\frac{\pi }{5}+{\sin }^2\frac{\pi }{5})$

$z_0{z}_1{z}_2{z}_3{z}_4$$=1$ since ${\cos }^2\theta +{\sin }^2\theta =1$