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Q.

Let, ‘ ’ be nth root of unity and z be a complex number such that  |zk|1k=0,1,....,n1 . Then  |z| can be___________

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a

1

b

0

c

2

d

3

answer is B.

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Detailed Solution

(zk)(z¯k¯)1

zz¯zk¯Z¯k+kk¯1

|z|2zk¯+z¯k

Now, sum these relations  |z|2×nz(k=0n1k)+z¯(k=0n1k)

|z|2n0|z|=0

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