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If the complex number $\frac{z - 1}{z + 1}$ is purely imaginary, then

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|z|=1

|z|<1

|z|>1

|z|≥2

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detailed solution

Correct option is A

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

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If the complex number $\frac{z - 1}{z + 1}$ is purely imaginary, then