Question

$\displaystyle\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=$

• A. $i$
• B. $-i$
• C. $2 i$
• D. $-2 i$

Answer 1

Answer: (A)

$ \sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7} $
$ =-i\left(\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}\right) $
$ =-i e^{i \frac{2 \pi k}{7}} $
$ \displaystyle\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=-\displaystyle\sum_{k=1}^6 e^{i \frac{2 \pi k}{7}} \ldots \text { (i) }$
$ \because Z^7-1=0 $
has roots $ Z=e^{i \frac{2 \pi k}{7}}, k=0,1,2,3,4,5,6 $
$ \therefore \displaystyle\sum_{k=0}^6 e^{i \frac{2 \pi k}{7}}=0$
$ \Rightarrow [\text { Sum of roots of unity is } 0] $
$\Rightarrow \displaystyle\sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}=0 t e^6=-1 \frac{2 \pi k}{7} $
$ =-i\left(\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}\right)$
$ =-i e^{i \frac{2 \pi k}{7}}$
$ \displaystyle\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=-\displaystyle\sum_{k=1}^6 e^{i \frac{2 \pi k}{7}} \ldots \text { (i) } $
$ \because Z^7-1=0$
has roots $Z=e^{i \frac{2 \pi k}{7}}, k=0,1,2,3,4,5,6$
$\therefore \displaystyle\sum_{k=0}^6 e^{i \frac{2 \pi k}{7}}=0[\text { Sum of roots of unity is } 0] $
$ \Rightarrow 1+\displaystyle\sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}=0$
$ \Rightarrow t \displaystyle\sum_{k=1}^6 e^{i \frac{2 \pi k}{7}}=-1 $
$ \text { From Eq. (i), }$
$\displaystyle\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=(-i)(-1)=i$