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Suppose that $$f$$ is a Mobius transformation such that $$f(1)=i$$, $$f(i)=-1,$$ and $$f(-1)=1$$. Find the value of $$f(-i)$$.

Jan 26, 2020

### 2+0 Answers

#1
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f(-i) = 4 + 2i.

Jan 27, 2020
#2
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Suppose that f is a Mobius transformation such that $$f(1)=i$$, $$f(i)=-1$$, and $$f(-1)=1$$.
Find the value of $$f(-i)$$.

Möbius transformation: $$w={\dfrac {az+b}{cz+d}}$$ of one complex variable z; here the coefficients a, b, c, d are complex number.

$$a=\det {\begin{pmatrix}z_{1}w_{1}&w_{1}&1\\z_{2}w_{2}&w_{2}&1\\z_{3}w_{3}&w_{3}&1\end{pmatrix}}\, \\ b=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&w_{1}\\z_{2}w_{2}&z_{2}&w_{2}\\z_{3}w_{3}&z_{3}&w_{3}\end{pmatrix}}\, \\ c=\det {\begin{pmatrix}z_{1}&w_{1}&1\\z_{2}&w_{2}&1\\z_{3}&w_{3}&1\end{pmatrix}}\, \\ d=\det {\begin{pmatrix}z_{1}w_{1}&z_{1}&1\\z_{2}w_{2}&z_{2}&1\\z_{3}w_{3}&z_{3}&1\end{pmatrix}}$$

$$\begin{array}{|rll|} \hline f(z)=w \\ \hline f(1)=i: & z_1=1 & w_1 =i \\ f(i)=-1: & z_2=i & w_2 = -1 \\ f(-1)=1: & z_3=-1 & w_3 = 1 \\ \hline \end{array}$$

$$\begin{array}{|lcll|} \hline a= \begin{vmatrix}i&i&1\\-i&-1&1\\-1&1&1\end{vmatrix}&=& -4i-2 \\\\ b= \begin{vmatrix}i&1&i\\-i&i&-1\\-1&-1&1\end{vmatrix}&=& -2 \\\\ c= \begin{vmatrix}1&i&1\\i&-1&1\\-1&1&1\end{vmatrix}&=& -2 \\\\ d= \begin{vmatrix}i&1&1\\-i&i&1\\-1&-1&1\end{vmatrix}&=& 4i-2 \\ \hline \end{array}$$

$$\begin{array}{|rll|} \hline f(z)=w &=& \dfrac {az+b}{cz+d} \\\\ \mathbf{f(z)} &=& \mathbf{\dfrac {(-4i-2)z-2}{(-2)z+(4i-2)}} \\ \hline \\ f(-i) &=& \dfrac {(-4i-2)(-i)-2}{(-2)(-i)+(4i-2)} \\\\ &=& \dfrac {4i^2+2i-2}{2i+4i-2} \quad | \quad i^2 = -1 \\\\ &=& \dfrac {-4+2i-2}{6i-2} \\\\ &=& \dfrac { 2i-6}{6i-2} \\\\ &=& \dfrac { 2(i-3)}{2(3i-1)} \\\\ &=& \dfrac { (i-3)}{ (3i-1)}*\dfrac{(3i+1)}{(3i+1)} \\\\ &=& \dfrac {3i^2+i-9i-3}{9i^2-1} \\\\ &=& \dfrac {-3+i-9i-3}{-9-1} \\\\ &=& \dfrac {-6-8i}{-10} \\\\ &=& \dfrac {6+8i}{10} \\\\ &=& \dfrac {6}{10} +\dfrac {8}{10}i \\\\ \mathbf{f(-i)} &=& \mathbf{0.6+0.8i} \\ \hline \end{array}$$

Jan 27, 2020