Let z1 and z2 be nth roots of unity which are ends of a line segment that subtend a right angle at the origin. Then n must be of the form [IIT Screening 2001; Karnataka 2002]
A
4k + 1
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B
4k + 2
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C
4k + 3
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D
4k
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Solution
The correct option is D4k 11/n=cos2rπn+isin2rπn Letz1=cos2rπn+isin2rπn andz2=cos2r2πn+isin2r2πn.
Then ∠Z1OZ2=amp(z1z2)=amp(z1)−amp(z2) =2(r1−r2)πn=π2 (Given) ∴n=4(r1−r2)=4× integer, so n is of the form 4 k.