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Question

The equation z10+(13z1)10=0 has 5 pairs of complex roots a1,b1,a2,b2,a3,b3,a4,b4,a5,b5. If each pair ai,bi are complex conjugates, then

A
5i=1(1ai+1bi)=130
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B
5i=1(1ai+1bi)=260
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C
5i=1(1aibi)=1700
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D
5i=1(1aibi)=850
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Solution

The correct option is D 5i=1(1aibi)=850
z10+(13z1)10=0z10[1+(131z)10]=0(131z)10=1131z=[cos(2m+1)π+isin(2m+1)π]1/10131z=ei(2m+1)π/101z=13ei(2m+1)π/10

For the complex roots, substituting m=0,1,2,,9
1a1=1z1=13ei(1π)/101b1=1z10=13ei(19π)/10=13ei(1π)/101a2=1z2=13ei(3π)/101b2=1z9=13ei(17π)/10=13ei(3π)/10 1a5=1z5=13ei(5π)/101b5=1z6=13ei(15π)/10=13ei(5π)/10

Now, 5i=1(1ai+1bi)
=13×102Re[ei(1π)/10+ei(3π)/10+ei(5π)/10+ei(7π)/10+ei(9π)/10]=1302[cosπ10+cos3π10+cos5π10+cos7π10+cos9π10]=1302[cosπ10+cos3π10+cosπ2cos3π10cosπ10]=130

Also, 1a1b1=(13ei(1π)/10)(13ei(1π)/10)
1a1b1=16913[ei(1π)/10+ei(1π)/10]+11a1b1=17026[cosπ10]

Now, 5i=1(1aibi)
=170×526[cosπ10+cos3π10+cos5π10+cos7π10+cos9π10]=850

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