$\displaystyle Let\ the\ bilinear\ transformation\ be\ \ w=\ \frac{az+b}{cz+d}\ \ \rightarrow{}\left(A\right) \\[3ex] \displaystyle (\ where\ a,b,c,d\ are\ complex\ constants\ and\ ad-bc\not=0) \\[3ex] \displaystyle Put\ z=0\ and\ w=-4i\ in\ (A) \\[2ex] \displaystyle \therefore{}-4i=\frac{b}{d} \\[2ex] \displaystyle b=\ -4id\rightarrow{}\left(B\right) \\[2ex] \displaystyle Put\ z=i,\ w=\infty{}\ in\ \left(A\right) \\[2ex] \displaystyle \infty{}=\frac{ai+b}{ci+d}\ \\[3ex] \displaystyle \therefore{}ci+d=0 \\[2ex] \displaystyle \therefore{}c=\ -\frac{d}{i} \\[2ex] \displaystyle c=\ di\rightarrow{}(C) \\[2ex] \displaystyle Put\ z=-2i\ ,\ w=0\ in\ \left(A\right) \\[2ex] \displaystyle 0=\frac{-2ai+b}{-2ci+d} \\[3ex] \displaystyle \therefore{}b=2ai \\[2ex] \displaystyle from\ \left(B\right),\ -4id=2ai\ \\[2ex] \displaystyle a=\ -\frac{4id}{2i}=\ -2d\rightarrow{}\left(D\right) \\[2ex] \textit{From (A),(B),(C),(D)}\\[2ex] \displaystyle w=\ \frac{az+b}{cz+d} \\[2ex] \displaystyle w=\ \frac{-2dz-4id}{diz+d} \\[2ex] \displaystyle \ w=-\frac{2z+4i}{iz+1} \\[2ex] Fixed\ \left(or\ invariant\right)points\ of\ the\ transformation\ w=f\left(z\right)are\ points\\[2ex] which\ are\ mapped\ onto\ themselves.\\[3ex] $

$\therefore{}w=f\left(z\right)= \displaystyle w=-\frac{2z+4i}{iz+1}=z \\[2ex] z\left(iz+1\right)=\ -2z-4i\\[2ex] \displaystyle iz^2+z+2z+4i=0 \\[2ex] \displaystyle iz^2+3z+4i=0 \\[2ex] {This \ is \ quadratic \ in `z' \ having \ 2 \ roots, \ which \ confirms \ the \ fact \\[1ex] that \ bilinear \ transformation \ has \ 2 \ fixed \ points.}\\[2ex] \displaystyle iz^2-z+4z+4i=0 \\[2ex] \displaystyle iz^2+i^2z+4z+4i=0 \\[2ex] \displaystyle iz\left(z+i\right)+4\left(z+i\right)=0 \\[2ex] \displaystyle \left(z+i\right)\left(iz+4\right)=0 \\[2ex] \displaystyle z+i=0\ ,\ iz+4=0\ \\[2ex] \displaystyle z=\ -i\ ,\ z=\ -\frac{4}{i} \\[2ex] \displaystyle z=-i, \\[3ex]4i\ are\ the\ fixed\ points\ of\ the\ transformation. \\[2ex] $