# Bilinear transformation $f ( z ) = 9 e ^ {i \theta } \frac { ( z - 1 ) } { ( z - 9 ) } , \theta \in [ 0,2 \pi ]$

$$\begin{array} { l } { \text { Consider the Bilinear transformation } f ( z ) = 9 e ^ {i \theta } \frac { ( z - 1 ) } { ( z - 9 ) } , \theta \in [ 0,2 \pi ] } \\ { \text { then which of the following is/are correct } } \\ { \text { (a) } |f \left( 3 e ^ { i \theta } \right) |= 3 , \forall \theta \in [ 0,2 \pi ] } \\ { \text { (b) } |f \left( 3 e ^ { i \theta } \right)| = 1 , \forall \theta \in [ 0,2 \pi ] } \\ { \text { (c) } |f \left( 3 e ^ { i \theta } \right)| = 2 , \forall \theta \in [ 0,2 \pi ] } \\ { \text { (d) } f ( z ) \text { is conformal on } \mathbb { C } - \{ 9 \} } \end{array}$$ $$\begin{array} { l } { \textbf { My attempt:- } } \\ { \text { (d) Correct, Since f(z) a mobious transformation } } \\ { \text { (c) and (b) Can be eliminated by taking \theta=2\pi } } \\ { \text { Clearly f(z) is a rotation by } \theta \text { degrees } } \\ { \text { and dialation by } 9 \text {, Then how } |f \left( 3 e ^ { i \theta } \right) |= 3 ? } \end{array}$$

• In your defining equation for $f$, what is $\theta$? $f$ is a function of $z$, is $\theta$ supposed to be a constant? Is it a 2nd variable, so it is actually $f(\theta, z)$ not $f(z)$? or should your function actually be $f(z) = 9\frac z{|z|}\frac {z-1}{z-9}$?. Unless $\theta$ is a constant (in which case using it again in the questions in an apparent different meaning is a Cardinal Sin of Mathematics), you are mistaken about $f$ being a Möbius transformation – Paul Sinclair Oct 10 at 0:30
• First of all you have to fix a $\theta$ – RIYASUDHEEN T. K Oct 10 at 1:20
• Then why do the first three questions have $\forall \theta \in [0,2\pi]$? If $\theta$ is a fixed constant set in the definition of $f$, then the $\forall$ clauses here don't make sense, as $\theta$ already has a set value. – Paul Sinclair Oct 10 at 16:33
• $|f(3e^{i \theta})|=|9e^{i \theta}\frac{3e^{i \theta}-1}{3e^{i \theta}-9}|$ , Then taking $3$ out side of the denominator to get $$|f(3e^{i \theta})|=|3e^{i \theta}\frac{3e^{i \theta}-1}{e^{i \theta}-3}|$$ $$=3|\frac{3e^{i \theta}-1}{e^{i \theta}-3}|$$ And since $|\frac{z}{\bar{z}}|=1$ answer is 3 and this is true for any fixed $\theta.$ – RIYASUDHEEN T. K Oct 11 at 7:02
• That has nothing at all to do with what I am talking about. Each of the questions (a) (b) (c) have attached to them "$\forall \theta \in [0,2\pi]$" That quite succinctly indicates that $\theta$ is not fixed here. but it contradicts the use of $\theta$ in the definition of $f(z)$. My suspicion is that the $\theta$ in the questions is actually supposed to be a separate variable than the $\theta$ in the definition (a practice that I keep seeing people do despite how obviously bad it is). I am trying to clarify that matter. – Paul Sinclair Oct 11 at 18:02