Find the image of the unit circle under the transformation $f(z)=\frac{z+1}{2z+1}$. How Do I approach these questions? Find the image of the unit circle under the transformation $f(z)=\frac{z+1}{2z+1}$. How Do I approach these questions?
I tried writing $z$ as $e^{i \phi}$, but I didn't know how to continue from there.
Thank you for any assistance! 
 A: You know $|z| = 1$. Let $A = \frac{z+1}{2z+1}$.
$$(2z+1)A = z + 1 \\
z(2A-1) = 1 - A\\
z = \frac{1-A}{2A-1}\\
1 = \frac{|1-A|}{|2A-1|}\\
|1-A| = |1 - 2A|\\
|1-A| = 2 \left|\frac{1}{2} - A \right|$$
The set of all such $A$ is always a circle.
Next, find two polar opposite points on the circle; that gives you the centre and radius.
A: HINT: This is an example of Möbius transformation. Try to decompose the transformation $z\mapsto\frac{z+1}{2z+1}$ into simpler transformations as indicated in the linked page and follow the unit circle through.

For a nice depiction of Möbius transformations, see here.
A: Let $C$ be the unit circle. Since $f$ is a Moebius transformation with no singularity on $C,$ $f(C)$ is a circle. Now $f$ is symmetric about the real axis, hence so is $f(C).$ It follows that $[f(-1),f(1)] = [0,2/3]$ is a diameter of $f(C).$ Therefore $f(C)$ is the circle $(x-1/3)^2 +y^2 = 1/9.$
A: i think you can use some of the nice properties of the moebius transformation to obtain the image of the unit circle without too much computation. i will use two properties: (a) circles transformed into circles (b) inverse points transform to inverse points
$-\frac 12$ and  $2$ are inverse points on the unit circle. they are transformed by $w = \frac{z+1}{2z+1}$ to the inverse points $\infty, \frac 13.$  therefore the center of the image circle is $\frac 13.$  
to find the radius we will transform the point $1$ on the unit cicle that goes to $\frac23$ therefore the radius is $\frac 13.$
so the image of $$|z| = 1 \to \Big|w- \frac 13\Big|= \frac 13. $$
