document.write( "Question 31235: What is the value of Summation(k from 1 to 6(sin(2*pi*k/7)+i*cos(2*pi*k/7)));
\n" ); document.write( "where k is a number taking the values 1 to 6;
\n" ); document.write( "pi is 22/7;
\n" ); document.write( "i is the square-root of (-1); \n" ); document.write( "
Algebra.Com's Answer #18003 by venugopalramana(3286)\"\" \"About 
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What is the value of Summation(k from 1 to 6(sin(2*pi*k/7)+i*cos(2*pi*k/7)));
\n" ); document.write( "where k is a number taking the values 1 to 6;
\n" ); document.write( "LET THE REQUIRED SUM =S
\n" ); document.write( "DEMOVIERS THEOREM STATES
\n" ); document.write( "{COS(2KPI)+iSIN(2KPI)}^(1/N)=COS(2KPI/N)+iSIN(2KPI/N)..THE N ROOTS OBTAINED BY PUTTING K=1,2,3,......N.
\n" ); document.write( "HERE IF WE PUT N=7 AND NOTE THAT.......COS(2KPI)+iSIN(2KPI)=1...SO...THE EQN.
\n" ); document.write( "Z^7=1...OR...Z^7-1=0...GIVES RAISE TO THE 7 VALUES OF 7 TH.ROOTS OF UNITY,WE GET
\n" ); document.write( "{COS(2KPI)+iSIN(2KPI)}^(1/7)=COS(2KPI/7)+iSIN(2KPI/7)..THE N ROOTS OBTAINED BY PUTTING K=1,2,3,.....7
\n" ); document.write( "IF WE CALL THEM Z1,Z2,Z3......Z7...THEN THEIR SUM IS
\n" ); document.write( "Z1+Z2+Z3+Z4+Z5+Z6+Z7= SUM OF 7 ROOTS OF UNITY OR THE EQN.....Z^7-1=0
\n" ); document.write( "BUT SUM OF ROOTS OF A POLYNOMIAL = -COEFFICIENT OF Z^6/COEFFICIENT OF Z^7=-0/1=0
\n" ); document.write( "HENCE
\n" ); document.write( "Z1+Z2+Z3+Z4+Z5+Z6+Z7= 0=S+Z7...AS ASKED FOR IN THE PROBLEM
\n" ); document.write( "Z7=COS(2*7PI/7)+iSIN(2*7PI/7)=COS(2PI)+iSIN(2PI)=1
\n" ); document.write( "HENCE
\n" ); document.write( "S+1=0
\n" ); document.write( "S=-1 \n" ); document.write( "
\n" );