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Q.

If the complex number z−1z+1 is purely imaginary, then

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a

|z|=1

b

|z|<1

c

|z|>1

d

|z|≥2

answer is A.

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Detailed Solution

z−1z+1== it where t∈R.⇒z¯−1z¯+1=−it ⇒z−1z+1+z¯−1z¯+1=0⇒(z−1)(z¯+1)+(z¯−1)(z+1)=0⇒2(zz¯−1)=0⇒|z|=1.

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