If the number $\frac{z - 2}{z + 2}$ is purely imaginary number, then modulus value of z satisfies

less than 2

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greater than 2

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lies between 2 and 2

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| z | = 2

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Solution

The correct option is D $| z | = 2$
Let $w = \frac{z - 2}{z + 2}$
Since, $w$ is purely imaginary
$∴ w + ¯ w = 0$
$\Rightarrow \frac{z - 2}{z + 2} + \frac{¯ z - 2}{¯ z + 2} = 0$
$\Rightarrow (z - 2) (¯ z + 2) + (z + 2) (¯ z - 2) = 0 \Rightarrow {| z |}^{2} + 2 z - 2 ¯ z - 4 + {| z |}^{2} - 4 + 2 ¯ z - 2 z - 4 = 0 \Rightarrow {| z |}^{2} = 4 \Rightarrow | z | = 2$
Hence, option 'D' is correct.

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If the number z−2z+2 is purely imaginary number, then modulus value of z satisfies

The correct option is D |z|=2Let w=z−2z+2Since, w is purely imaginary∴w+¯w=0⇒z−2z+2+¯z−2¯z+2=0⇒(z−2)(¯z+2)+(z+2)(¯z−2)=0⇒|z|2+2z−2¯z−4+|z|2−4+2¯z−2z−4=0⇒|z|2=4⇒|z|=2Hence, option 'D' is correct.

If the number $\frac{z - 2}{z + 2}$ is purely imaginary number, then modulus value of z satisfies