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

Question

\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}