Answer to Question #296319 in Complex Analysis for Lucky

Question #296319

Find the billianear transformation which maps the points z=0,1,∞ on to the points w=0,i,2i.


1
Expert's answer
2022-02-15T17:18:00-0500

"z_{1}=\\infty, z_{2}=i, z_{3}=0 \\& w_{1}=0, w_{2}=i, w_{3}=\\infty"

Let the transformation be,

"\\begin{gathered}\n\n\\frac{\\left(w-w_{1}\\right)\\left(w_{2}-w_{3}\\right)}{\\left(w-w_{3}\\right)\\left(w_{2}-w_{1}\\right)}=\\frac{\\left(z-z_{1}\\right)\\left(z_{2}-z_{3}\\right)}{\\left(z-z_{3}\\right)\\left(z_{2}-z_{1}\\right)} \\\\\n\n\\frac{\\left(w-w_{1}\\right) w_{3}\\left(\\frac{w_{2}}{w_{3}}-1\\right)}{w_{3}\\left(\\frac{w}{w_{3}}-1\\right)\\left(w_{2}-w_{1}\\right)}=\\frac{z_{1}\\left(\\frac{z}{z_{1}}-1\\right)\\left(z_{2}-z_{3}\\right)}{\\left(z-z_{3}\\right) z_{1}\\left(\\frac{z_{2}}{z_{1}}-1\\right)} \\\\\n\n\\frac{\\left(w-w_{1}\\right)\\left(\\frac{w_{2}}{w_{3}}-1\\right)}{\\left(\\frac{w}{w_{3}}-1\\right)\\left(w_{2}-w_{1}\\right)}=\\frac{\\left(\\frac{z}{z_{1}}-1\\right)\\left(z_{2}-z_{3}\\right)}{\\left(z-z_{3}\\right)\\left(\\frac{z_{2}}{z_{1}}-1\\right)} \\\\\n\n\\frac{(w-0)(0-1)}{(0-1)(i-0)}=\\frac{(0-1)(i-0)}{(z-0)(0-1)} \\\\\n\n\\frac{w}{i}=\\frac{i}{z} \\\\\n\nw=i^{2} \\\\\n\nw=\\frac{-1}{z}\n\n\\end{gathered}"

 


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