Which one of the following is true? 
The image of the half plane $\text {Re} (z) + \text {Im} (z) >0$ under the map $w  = \frac {z-1} {z+i}$ is given by
$(\text A)$ $|w| > 1$
$(\text B)$ $|w| < 1$

Let $z = x + iy$ with $x + y >0.$ Then I found that $$|w| = \frac {\sqrt {\left (x^2 + y^2 - x + y \right )^2 + \left (y - x + 1 \right)^2}} {x^2 + \left (y + 1 \right )^2}.$$
By taking $z = 1$ we have $|w| = 0 <1.$ So I believe that $(\text B)$ is the correct option. So we need only to prove that for any $x,y \in \Bbb R$ with $x + y > 0$ $$\sqrt {\left (x^2 + y^2 - x + y \right )^2 + \left (y - x + 1 \right)^2} < x^2 + \left (y + 1 \right )^2.$$
But how do I show that? Any help in this regard will be highly appreciated.
Thank you very much for your valuable time.
 A: Using $|z+i|^2=|z|^2+1+2\Re{(-iz)}=|z|^2+1+2\Im z$ and $|z-1|^2=|z|^2+1-2\Re z$, we get
$|z+i|^2-|z+1|^2=2(\Re z + \Im z) >0$ in our case, so $|z+i|^2>|z-1|^2$ or $|w|<1$ since obviously $z \ne -i$ and we can divide the inequality. Done! 
A: To continue from where you got up to, you could have tried
$$\begin{align}&\sqrt{\left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}}<x^{2}+\left(y+1\right)^{2}
\\
&\iff \left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}-\left(x^{2}+\left(y+1\right)^{2}\right)^{2}<0
\\
&\iff \left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{2}+y^{2}-x+y\right)^{2}+\left(y-x+1\right)^{2}-\left(x^{2}+\left(y+1\right)^{2}\right)^{2}}{x^{2}+\left(y+1\right)^{2}}\right)<0\tag{*}
\\
&\iff \tiny\left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{4}+2x^{2}y^{2}+y^{4}-2x^{3}+2x^{2}y-2xy^{2}+2y^{3}+2x^{2}-4xy+2y^{2}-2x+2y+1\right)-\left(x^{4}+2x^{2}y^{2}+y^{4}+4x^{2}y+4y^{3}+2x^{2}+6y^{2}+4y+1\right)}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff -2\left(x^{2}+\left(y+1\right)^{2}\right)\left(\frac{\left(x^{3}+xy^{2}+2xy+x\right)+\left(x^{2}y+y^{3}+2y^{2}+y\right)}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff -2\left(x^{2}+\left(y+1\right)^{2}\right)\left(x\cdot\frac{x^{2}+\left(y+1\right)^{2}}{x^{2}+\left(y+1\right)^{2}}+y\cdot\frac{x^{2}+\left(y+1\right)^{2}}{x^{2}+\left(y+1\right)^{2}}\right)<0
\\
&\iff {-2\left(x^{2}+(y+1)^{2}\right)\left(x+y\right)<0}
\\
&\iff x+y>0\quad \square
\end{align}$$
where, in $(*)$, we use the ansatz that the LHS is divisible by $\left(x^{2}+(y+1)^{2}\right)$.
