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.

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 $\left(x^{2}+(y+1)^{2}\right)$.
