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$z$ is a complex number, such that $$\frac{z-1}{z+1}$$ is purely imaginary.

Then what would be the absolute value of $z$?

Options were given as follows:

  • $|z|=0$
  • $|z|=1$
  • $|z|>1$
  • $|z|<1$
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    set $z=x+iy$... – 2017-01-21
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    @MyGlasses Yeah. I did that. And I ended up getting some equation in terms of difference of x^2 and y^2 – 2017-01-21

5 Answers 5

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$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\times\frac{\bar{z}+1}{\bar{z}+1}=\frac{z\bar{z}+z-\bar{z}-1}{|z+1|^2}=\frac{|z|^2+2i{\bf Im}(z)-1}{|z+1|^2}$$ Thia expression is purely imaginary, so $|z|^2-1=0$ or $|z|=1$.

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Geometrically $z-1$ is the vector with origin in the Argand plane at $z_1=1$ and end at $z$. Similarly $z+1$ has origin at $z_2=-1$ and end at $z$. The argument of ${z-1\over z+1}$ is the angle between those two vectors. When the ratio is purely imaginary the angle is ${\pi\over 2}$. So in the Argand plane $z$ is on the circle of diameter $z_1z_2$, so the answer is $|z|=1$

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    Wow. Nice explanation. A bit different from the conventional approach. Thanks – 2017-01-21
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    Approaching visually (geometrically) complex analysis has helped me a lot – 2017-01-21
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You are given that $$\frac{z-1}{z+1}=ci\text{ for some real $c$.}$$ Let's solve for $z$, then we'll find $|z|$.

There are a couple of ways to solve for $z$. Let's just directly isolate $z$.

\begin{align} \frac{z-1}{z+1}&=ci\\ z-1&=(z+1)ci\\ z-1&=zci+ci\\ z-zci&=1+ci\\ z(1-ci)&=1+ci\\ z&=\frac{1+ci}{1-ci} \end{align}

Rationalizing the denominator results in $$z=\frac{1-c^2}{1+c^2}+\frac{2c}{1+c^2}\ .$$ Fighting through the algebra, we find that $|z|=1$.

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    @WorldSEnder Good point! I've updated my answer. – 2017-01-21
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set $$z=a+bi$$ then we get $$\frac{a-1+bi}{a+1+bi}=ci$$ now you can calculate $$a,b,c$$

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use the fact that

$z+\bar z= 2Re(z)$

now, let $w=$ $ z-1\over z+1$

since $w$ is purely imaginary, this implies $w+\bar w=0 $

so $ z-1\over z+1$ $+$ $ \bar z-1\over \bar z+1$ $= 0$

$z\bar z+z-\bar z-1+z\bar z+\bar z-z-1 \over(z+1)(\bar z+1)$ $=0$

$z\bar z-1=0$

$|z|^2-1=0$

$ |z|=1$