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Question

If Sk=sin(2πkn+1)icos(2πkn+1) and Pk=cos(π2k)+isin(π2k),
Let m=nk=1Sk+k=1Pk, then the value of 2m2 is

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Solution

Sk=sin(2πkn+1)icos(2πkn+1)
=i[cos(2πkn+1)+isin(2πkn+1)]
=iei2πkn+1
nk=1Sk=ink=1zk, [z=ei2πn+1]
=i(z+z2+z3++zn)
=iz(zn1)z1
=i(zn+1zz1)
Now, zn+1=e(i2πn+1)(n+1)=ei2π=1
nk=1Sk=i(1z)z1=i

Pk=cos(π2k)+isin(π2k)=eiπ2k
k=1Pk=k=1eiπ2k
=eiπ2eiπ22eiπ23
=exp[iπ(12+122+123+)]
=eiπ=1
nk=1Sk+k=1Pk=|i1|=2

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