Let the bilinear transformation be $w=\dfrac {az+b}{cz+d} \ \ \to (1)$ (where a,b,c,d are complex constant & ad-bc≠0

$w=\dfrac{z(a+b/z)}{z(c+d/z)} = \dfrac {a+b/z}{c+d/z}$

Put z=∞ (∴1/z=0) and w=i

$\therefore i = \dfrac {a+0}{c+0} $

$\therefore ic =a\to (2)$

Put z=-1 and w=-1 in (1)

$-i = \dfrac {a(-1) +b}{c(-1) +d}$

$\therefore i\ c -i \ d = -a+b $

$\therefore i\ c - i \ d = - i \ c + b \ (From \ 2)$

$\therefore 2i \ c- i\ d = b \to (3)$

Put z=1 and w=-1 in (1)

$\therefore -1 = \dfrac {a(1)+b}{c(1)+d}$

$\therefore -c - d = a+ b \ \to (4)$

Substitute (2) and (3) in (4)

$\therefore -c - d = ic + 2i \ c - i \ d$

$\therefore i \ d - d = i \ c + 2i \ c + c$

$\therefore d(i-1)=3i \ c+c$

$\therefore d=\dfrac {(3i+1)c}{i=1}$

$\therefore d=(1-2i) \ c\to (5)$

Substitute (5) in (3)

$\therefore b=2i \ c-i (1-2i) \ c=(-2+i)c$

$\therefore w=\dfrac {z(ic)+(-2+i)c}{c \ z+(1-2i)c}$

$\therefore w=\dfrac {iz - 2 +i}{z+1-2i}$

is the required bilinear transformation.