Determine maximum of $|f(z)| \: \: \mbox{for}\: z \in \{z \in \mathbb{C} : |z| \leq 1\} $ Let $f(z) = \frac{z+3}{z-3}$. How to calculate the maximum of $|f(z)|$ for $z \in \{z \in \mathbb{C} : |z| \leq 1\}$? I've tried $z = \exp(it) \:\: t \in \mathbb{R}$, but I wasn't able to get a solution this way. Any hints?
 A: As $w=\frac{z+3}{z-3}$. This on simplification gives you $z=\frac{3w+3}{w-1}$.
You need the image of $|z|\le 1$ for which you can just plug in the above value of $z$ here:
$\frac{3w+3}{w-1}\le 1\implies |3w+3|\le |w-1|$.
Plug $w=u+iv$ and simplify to get the circle 
$u^2+v^2+\frac{9}{4}u+\frac{1}{4}v+1=0$ with center at $(-9/8,-1/8)$ and radius $\sqrt{18}/8$.
A: You are lucky here because of the choice of $f$. First notice that as $|z|\leq 1$ doesn't have any singularities of $f$, the image of $|z|=1$ under $f$ will be a circle. The $max_{|z|\leq1}\{|f(z)|\}$ will occur if $f(z)$ is on the boundary of the image and is farthest from the origin. By, previous observation we observe that we need to maximize $$\frac{|z+3|}{|z-3|}\quad given \quad  |z| =1$$
Now, you are lucky because maximum value of $|z+3|$ and minimum value of $|z-3|$ occurs at same value of $z$; i.e. at $z=1$. Therefore we get $$max_{|z|\leq1}\{|f(z)|\} = 2$$ 
PS: Whenever you have a question on mobius transformation, draw the picture first to understand what is happening. :)
