Find the absolute value of $z$ for the condition $z$ 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: 


*

*$|z|=0$

*$|z|=1$

*$|z|>1$

*$|z|<1$

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

A: set $$z=a+bi$$ then we get
$$\frac{a-1+bi}{a+1+bi}=ci$$
now you can calculate $$a,b,c$$
A: 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$
