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WBJEE 2012 Math Solved Paper

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Question : 33 of 80

Marks: +1, -0

If

\frac{z-1}{z+1}

is purely imaginary, then

$|z| = \frac{1}{2}$
$|z| = 1$
$|z| = 2$
$|z| = 3$

Solution:

Let

\omega = \frac{z-1}{z+1}

Then, using componendo and dividendo we set

z = \frac{1+\omega}{1-\omega}

\Rightarrow |z| = \left| \frac{\omega+1}{\omega-1} \right|

Put

|z| = 1

\Rightarrow |\omega+1| = |\omega-1|

...(i)

Let

\omega = u+iv

Then

|\omega+1| = |u+iv+1|

= \left| (u+1)+iv \right| = \sqrt{(u+1)^{2}+v^{2}}

and

\left| \dot{\omega} - 1 \right| = \sqrt{(u-1)^{2}+v^{2}}

∴ From Eq. (i)

\sqrt{(u+1)^{2}+v^{2}} = \sqrt{(u-1)^{2}+v^{2}}

\Rightarrow (u+1)^{2}+v^{2} = (v-1)^{2}+v^{2}

\Rightarrow u = 0

\therefore \omega = \frac{z-1}{z+1}

is a pure imaginary number

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