The equation z10+(13z−1)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
5∑i=1(1ai+1bi)=130
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B
5∑i=1(1ai+1bi)=260
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C
5∑i=1(1aibi)=1700
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D
5∑i=1(1aibi)=850
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Solution
The correct option is D5∑i=1(1aibi)=850 z10+(13z−1)10=0⇒z10[1+(13−1z)10]=0⇒(13−1z)10=−1⇒13−1z=[cos(2m+1)π+isin(2m+1)π]1/10⇒13−1z=ei(2m+1)π/10∴1z=13−ei(2m+1)π/10
For the complex roots, substituting m=0,1,2,…,9 1a1=1z1=13−ei(1π)/101b1=1z10=13−ei(19π)/10=13−e−i(1π)/101a2=1z2=13−ei(3π)/101b2=1z9=13−ei(17π)/10=13−e−i(3π)/10⋮1a5=1z5=13−ei(5π)/101b5=1z6=13−ei(15π)/10=13−e−i(5π)/10