The value of displaystyle∑r=116( sin(2rπ/17)+i cos (2rπ/17)) is

Question

The value of displaystyle∑r=116( sin(2rπ/17)+i cos (2rπ/17)) is (A) 1 (B) i (C) - i (D) - 1

Tardigrade Admin · Accepted Answer

We have displaystyle∑r=116 (sin (2r π/17)+i cos (2r π/17)) =i displaystyle∑r=116 (cos (2r π/17)+i sin (2r π/17)) =i(z+z2+......+z16) [where z= cos (2 π/17)-i sin (2 π/17)] =i (z(1-z16)/1-z)=(i(z-z17)/1-z)=(i(z-1)/1-z)=i [∵ z17=1]

Q. The value of $r = 1 \sum 16 (sin \frac{2 r π}{17} + i cos \frac{2 r π}{17})$ is

$1$

$i$

$- i$

$- 1$

Q. The value of r=1∑16​(sin172rπ​+icos172rπ​) is

1

i

−i

−1

We have r=1∑16​ (sin172rπ​+icos172rπ​) =ir=1∑16​ (cos172rπ​+isin172rπ​) =i(z+z2+......+z16) [where z=cos172π​−isin172π​] =i1−zz(1−z16)​=1−zi(z−z17)​=1−zi(z−1)​=i [∵z17=1]

Q. The value of $r = 1 \sum 16 (sin \frac{2 r π}{17} + i cos \frac{2 r π}{17})$ is

$1$

$i$

$- i$

$- 1$