If Zr = sin (2 π r/11) - i cos (2 π r/11) then displaystyle∑10r = 0 Zr =

Question

If Zr = sin (2 π r/11) - i cos (2 π r/11) then displaystyle∑10r = 0 Zr =  (A) -1 (B) 0 (C) i (D) -i

Tardigrade Admin · Accepted Answer

We have, Zr = sin (2 π r/11)-i cos (2 π r/11) =-i( cos (2 π r/11)+i sin (2 π r/11)) =-i e(i 2 π r/11) displaystyle∑r=010 Zr =-i displaystyle∑r=010 e(i 2 π r/11) =-i × 0=0

Q. If $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$ then $r = 0 \sum 10 Z_{r} =$

-1

0

i

-i

We have, $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$
$= - i (cos \frac{2 π r}{11} + i sin \frac{2 π r}{11})$
$= - i e^{\frac{i 2 π r}{11}}$
$r = 0 \sum 10 Z_{r} = - i r = 0 \sum 10 e^{\frac{i 2 π r}{11}}$
$= - i \times 0 = 0$

Q. If Zr​=sin112πr​−icos112πr​ then r=0∑10​Zr​=

-1

0

i

-i

We have, Zr​=sin112πr​−icos112πr​ =−i(cos112πr​+isin112πr​) =−ie11i2πr​ r=0∑10​Zr​=−ir=0∑10​e11i2πr​ =−i×0=0

Q. If $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$ then $r = 0 \sum 10 Z_{r} =$

We have, $Z_{r} = sin \frac{2 π r}{11} - i cos \frac{2 π r}{11}$
$= - i (cos \frac{2 π r}{11} + i sin \frac{2 π r}{11})$
$= - i e^{\frac{i 2 π r}{11}}$
$r = 0 \sum 10 Z_{r} = - i r = 0 \sum 10 e^{\frac{i 2 π r}{11}}$
$= - i \times 0 = 0$