Let z1 and z2be nth roots of unity which subtend a right angle at the origin, then n must be of the form (where, k is an integer)
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 Since, argz1z2=π2⇒z1z2=cosπ2+isinπ2=i∴zn1zn2=(i)n⇒in=1⇒n=4k Alternate Solution argargz2z1=π2∴z2z1=∣∣z2z1∣∣eiπ2⇒z2z1=i⇒(z2z1)n=in ∴z1 and z2 are nth roots of unity. zn1=zn2=1⇒(z2z1)n=1⇒in=1⇒n=4k, where k is an integer.