Question

Determine the bilinear transformation which maps z1=0,z2=1,z3=infinite onto w1=i,w2=-1,w3=-i respectively.​

Answer 1

Answer:

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Step-by-step explanation:

Find the bilinear transformation that maps distinct points z1, z2, z3 onto the points w1 = 0, w2 = 1, w3 = ∞. - Slader.

Answer 2

Bilinear transformation: A transformation $w=\frac{az+b}{cz+d}$  where,

• a, b, c, d are real/complex constants
• such that ad-bc ≠ 0

A bilinear transformation preserves the cross ratio of four points.

That is,   $\frac{(w-w_{1})(w_{2}-w_{3})}{(w_{1}-w_{2})(w_{3}-w)}=\frac{(z-z_{1})(z_{2}-z_{3})}{(z_{1}-z_{2})(z_{3}-z)}$  - - - - - (1)

Given, $z1=0, z2=1, z3= \infty,$

          $w1=i,w2=-1,w3=-i$

 We have,         $\bf\frac{(w-w_{1})(w_{2}-w_{3})}{(w_{1}-w_{2})(w_{3}-w)}=\frac{(z-z_{1})(z_{2}-z_{3})}{(z_{1}-z_{2})(z_{3}-z)}$- - - - (2)

We know that the value of $z_{3}=\infty$, to avoid infinity value let us take out $z_{3}$ common in numerator and denominator in (2)

$\bf\frac{(w-w_{1})(w_{2}-w_{3})}{(w_{1}-w_{2})(w_{3}-w)}=\frac{(z-z_{1})z_{3}(\frac{z_{2}}{z_{3}}-1)}{(z_{1}-z_{2})z_{3}(1-\frac{z}{z_{3}})}$

Clearly $z_{3}$ on the RHS cancels out,

We have,

        (w,i,-1,i) = (z,0,1,∞)

Substituting these values in (1), we get,

          $\bf\frac{(w-i)(-1-(-i))}{(i-(-1))(-i-w)}=\frac{(z-0)(\frac{1}{\infty}-1)}{(0-1)(1-\frac{z}{\infty})} \\\\\\ \frac{(w-i)(-1+i)}{(i+1)(-i-w)}==\frac{z(-1)}{-1(1-0)}\\\\\\ \frac{(w-i)(i-1)}{(i+1)(-i-w)}=z}$- - - -(3)

Let us simplify LHS further,

$\bf\frac{(w-i)(i-1)}{(i+1)(-i-w)}=\frac{(w-i)(i-1)\times (i+1)}{(i+1)(-i-w)\times(i+1}$    

 $using\,\,(a+b)^2=a^2+2ab+b^2\,\,for\,\,(i+1)^2\,\, where \,\,a=i\,\,&\,\,b=1\\\\ we\,\,get\,\,(i+1)^2=i^2+2i+1=-1+2i+1=2i\\\\Let\,\,us\,\,use\,\,(a+b)(a-b)=a^2-b^2\,\,for\, (i+1)(i-1)\,\,\\\\(1+i)(1-i)=i^2-1^2=-1-1=-2$

               

                          $\bf=\frac{(w-i)(i-1)\times (i+1)}{(i+1)(-i-w)\times(i+1)}\\\\=\frac{(-2)(w-i)}{2i\,(-i-w)}$

        $\bf\frac{(w-i)(i-1)}{(i+1)(-i-w)}=\frac{2(w-i)}{2i\,(i+w)}$

∴ (3) becomes,

$\frac{(z-0)(\frac{1}{\infty}-1)}{(0-1)(1-\frac{z}{\infty})}=\frac{z(-1)}{-1(1-0)}$