Proving $\frac{z}{1+tz}$ maps upper half plane to itself Prove $T(z)=\frac{z}{1+tz}$ maps upper half plane to itself, where $t>0$.
I need to prove the imaginary part of $\frac{z}{1+tz}$ is positive. Here is my attempt:
Let $z = a+bi$, where $b>0$, $a \in \mathbb{R}$ and $t>0$.
$$\frac{a+bi}{1+t(a+bi)}=\frac{a+bi}{1+ta+tbi}=\frac{a+bi}{1+ta+tbi}\frac{1+ta-tbi}{1+ta-tbi}=\frac{a^2t+a+b^2t+ib}{a^2t^2+2at+b^2t^2+1}=\frac{a^2t+a+b^2t}{a^2t^2+2at+b^2t^2+1}+\frac{ib}{a^2t^2+2at+b^2t^2+1}$$
But that $2at$ in the denominator is not guaranteed to be positive, so I don´t know how to proceed.
Thanks.
 A: Complete the square:
$$a^2t^2+2at+b^2t^2+1=(at+1)^2+b^2t^2>0.$$
Else write $q(t)=(a^2+b^2)t^2+2at+1$ and consider the discriminant 
$$B^2-4AC=(2a)^2-4(a^2+b^2)(1)=-4b^2<0,$$
which implies $q$ has no real roots. It is a '$\bigcup$'-quadratic and if it has no roots we must have $q>0$.
Alternatively write
$$\frac{z}{1+tz}=\frac{z}{1+tz}\cdot \frac{1+t\overline{z}}{1+t\overline{z}}=\frac{z+t|z|^2}{|1+tz|^2}=\frac{z}{|1+tz|^2}+\underbrace{\frac{t|z|^2}{|1+tz|^2}}_{\in\mathbb{R}},$$
which has imaginary part 
$$\frac{\Im(z)}{|1+tz|^2}>0.$$
A geometric answer looks as follows: the argument of $tz$ is the same as the argument of $z$, $\arg z$.

Adding one to $tz$ translates it one unit in the complex plane to the right. This decreases the argument of $tz$, and means that $\arg (1+tz)$ is less than $\arg tz$, and so less than $\arg z$. The difference between them is $\alpha>0$:

Dividing by $1+tz$ involves a rotation through an angle $-\arg(1+tz)=-\theta$. This rotation is by an angle less than $\arg(z)$. This means that after the rotation,  by $-\arg(1+tz)$, by $r_{-\arg(1+tz)}$, $r_{-\arg(1+tz)}(z)=:w$ is still in the upper half plane (because it's argument is $\alpha$ such that $\arg z>\alpha>0$:

There is only a scaling of $w$ by $\displaystyle \frac{1}{|1+tz|}$ to find $\displaystyle \frac{z}{1+tz}$ and this doesn't 'push' $w$ outside the upper half plane (picture has wrong scale factor):

A: The denominator is $(1+ta+tib)(1+ta-tib)$, which is a nonzero
complex number multiplied by its conjugate. A nonzero complex number multiplied
by its conjugate is a positive real number.
A: $w=\frac{z}{1+tz}\implies z=\frac{w}{1-tw}$.
Now the upper half $z-$plane is $y>0$ or $\frac{z-\overline z}{2i}>0$
$\implies \frac{w}{1-tw}-\frac{\overline w}{1-t\overline w}>0$
$w-\overline w>0$ or $Im(w)>0$
