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Question

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 D 4k
11/n=cos2rπn+i sin2rπn
Let z1=cos2rπn+i sin2rπn
and z2=cos2r2πn+i sin2r2πn.
Then Z1OZ2=amp(z1z2)=amp(z1)amp(z2)
=2(r1r2)πn=π2 (Given)
n=4(r1r2)=4× integer, so n is of the form 4 k.

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