If $ \frac{1-z}{1+z} $ is purely imaginary, then what will be $ \vert z \vert $? Assuming $ z $ a non-zero complex number; the ratio $ \frac{1-z}{1+z} $ is purely imaginary. Then what will be $ \vert z \vert $ ? 
I can't figure out a clear, simple way to figure out its real part. How to do it?
 A: IMO problems like this should be solved without using imaginary/real parts.
You were given that something is pure imaginary. Therefore that something is negated under complex conjugation. IOW
$$
\frac{1-z}{1+z}=-\overline{\left(\frac{1-z}{1+z}\right)}=-\frac{1-\overline{z}}{1+\overline{z}}.
$$
Thus
$$
(1-z)(1+\overline{z})=-(1-\overline{z})(1+z).
$$
Expanding both sides you see that the terms $-z+\overline{z}$ cancel, and, recalling that $z\overline{z}=|z|^2$, you are left with the equation
$$1-|z|^2=-(1-|z|^2).$$
Leaving this to you.
A: Say $${1-z\over 1+z} =ib\;\;\;\Longrightarrow \;\;\;z = {1-bi \over 1+bi}$$
so $$|z| = {|1-bi|\over |1+bi|} = 1$$
A: Hint: set $z = a + ib$. You then have, 
$$\frac{1-a-ib}{1+a+ib}$$
Start by multiplying by $$\frac{1-a-ib}{1-a-ib}$$. 
A: Hint. One may observe that
$$
\frac{1-z}{1+z}=\frac{(1-z)(1+\bar{z})}{(1+z)(1+\bar{z})}=\frac{1-|z|^2+2i\text{Im}z}{\left|1+z\right|^2}.
$$
A: Just to give a different argument, any linear fractional transformation $f(z)={az+b\over cz+d}$ (with $ad-bc\not=0$) takes circles to circles, where straight lines qualify as circles (of zero curvature).  For $f(z)={1-z\over1+z}$, we have $f(1)=0$ and $f(\pm i)=\mp i$.  Thus the (unique) circle determined by the points $1$, $i$, and $-i$ maps to the (unique) circle determined by the points $0$, $-i$, and $i$.  The former is the unit circle, and the latter is the imaginary axis.  So $f(z)={1-z\over1+z}$ is imaginary if and only if $|z|=1$ (except for $z=-1$, unless you want to think of $f(-1)=i\infty$ as imaginary).
The algebra, of course, has simply been swept under the rug of the general theorem about linear fractional transformations taking circles to circles.
A: Hint. Write $z=x+yi$ and expand the fraction by $1-(x+yi)$
A: Hint
Let $z = a+bi$, then
$$\frac{1-a-bi}{1+a+bi} = \frac{(1-a-bi)(1+a-bi)}{(1+a+bi)(1+a-bi)} = \frac{1+a-bi-a-a^2+abi-bi-abi+b^2}{(1+a)^2-b^2}=\frac{1-2bi-a^2+b^2}{1+a+a^2-b^2}=ci$$
for some $c \in \mathbb R$. What requirements on $a,b$ must be true for the real part to equal 0?
A: If $(1-z)/(1+z)$ is purely imaginary, then we can write
$$\frac{1-z}{1+z} = ai \qquad \exists\,a\in\mathbb R$$
solving this equation for $z$ gives $$ z = \frac{1-ai}{1+ai} = \frac{1-a^2}{1+a^2} -\frac{2a}{1+a^2}i,$$
so $$|z| = \sqrt{\left(\frac{1-a^2}{1+a^2}\right)^2 + \left(\frac{2a}{1+a^2}\right)^2},$$
which after a bit of algebraic simplification gives $|z| = 1$.
