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Put z=x+iy
w= \frac{x-1+iy}{x+iy+1}=  \frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}= \frac{2+iy}{2x+2}= \frac{1}{x+1}+ \frac{2yi}{2x+2}
Re(w)= \frac{1}{x+1}
in terms of z
x=(z+z')/2  where z'=x-iy=1/z
Re(w)= \frac{2}{z+z'+2}

1 5 1
Put z=cos α+i sinα
then solve for w
u can get
Re(z)=1/(cos α +1)
0 0 0
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