Find u(x,y) and v(x,y) for w=(1-z)/(1+z)

Find $u(x,y)$ and $v(x,y)$ for $\ w=\dfrac{1-z}{1+z}$.

Step 1 · Substitute and multiply by the conjugate

$$ w=\frac{(1-x)-iy}{(1+x)+iy}\cdot\frac{(1+x)-iy}{(1+x)-iy}=\frac{[(1-x)-iy][(1+x)-iy]}{(1+x)^2+y^2} $$

Step 2 · Expand the numerator

$$ [(1-x)-iy][(1+x)-iy]=(1-x)(1+x)-iy(1-x)-iy(1+x)+(iy)^2 $$

$$ =(1-x^2)-iy\big[(1-x)+(1+x)\big]-y^2=(1-x^2-y^2)-2iy $$

Step 3 · Read off $u$ and $v$

$$ w=\frac{(1-x^2-y^2)-2iy}{(1+x)^2+y^2} $$

$$ \boxed{u(x,y)=\frac{1-x^2-y^2}{(1+x)^2+y^2}, \qquad v(x,y)=\frac{-2y}{(1+x)^2+y^2}} $$

Verification, $z=0$

$u(0,0)=\dfrac{1-0-0}{1+0}=1,\ v(0,0)=\dfrac0{1}=0$, so $w=1$; direct check: $\dfrac{1-0}{1+0}=1$ ✓.

Verification, $z=i$

$u(0,1)=\dfrac{1-0-1}{1+1}=0,\ v(0,1)=\dfrac{-2}{2}=-1$, so $w=-i$; direct check: $\dfrac{1-i}{1+i}=\dfrac{(1-i)^2}{(1+i)(1-i)}=\dfrac{-2i}2=-i$ ✓.