IIT JEE , AIEEE Forum - Mathematics , Algebra

rtiit

8

sigma [sin(2r*pie/9) + icos(2r*pie/9)] =???

r=1

plz give detailed solution

elastiboysai

-1.
the summation includes the roots of the eqn z^9=1 all but for 1 itself.
sum of roots=0
so gvn sum =-1.

rtiit

no bt ans given here is i

sunip_the_mini_mastermind

yes the ans is i.
_{[r=1 ]}

^{[ 8]} sin (2r

/9)+icos(2r

/9)
=_{[r=1 ]}

^{[ 8]}-i² sin (2r

/9)+i cos(2r

/9)
=-i_{[r=1 ]}

^{[8 ]} i sin (2r

/9)+cos(2r

/9)
=-i_{[r=1 ]}

^{[ 8]} e^i2rpi/9
=-i_{[r=0 ]}

^{[ 8]} e^i2rpi/9-1
=-i(sum of n powers of unity -1) {sum of n powers of unity =0)
=i

rtiit