Let $ f:\Delta \mapsto \Delta $ be an analytic and bijective mapping. Let $ f:\Delta \mapsto \Delta $ be an analytic and bijective mapping. My question is whether this implies $ f (z)=kz $ for some $ k \in\mathbb {C} $ such that $| k|=1 $. Here, $\Delta :=\{z\in \mathbb { C}: |z|<1\} $.
 A: Not in general. Take any $\zeta\in\Delta$ and any $\alpha\in\Bbb R$. Then $$f(z)=e^{i\alpha}\frac{z-\zeta}{\overline\zeta z-1}\tag{$\clubsuit$}$$ is analytic in $\Delta,$ and maps $\Delta\to\Delta$ bijectively.

Development: Let's find the family of linear fractional transformations mapping $\Delta$ onto $\Delta$. (Linear fractional transformations are always injective.)
Take any such linear fractional transformation $$f(z)=\frac{az+b}{cz+d}$$ with $ad-bc\ne 0$. Note that if $\zeta\in\Delta$ is the unique point that $f$ maps to $0$, then $b=-a\zeta,$ so $az+b=a(z-\zeta)$.
A nice property of linear fractional maps is that given any circle or line in the plane, the linear fractional map maps points symmetric about that circle or line to points symmetric about the image of that circle or line. Another nice one is that the boundaries of disks will be mapped to the boundaries of the disks' images. In this case, that means that $f$ maps the unit circle to itself, so points symmetric about the unit circle will be mapped to points symmetric about the unit circle. In general, we say that points $z_1,z_2$ are symmetric about the circle $|z-z_0|=R$ if $(z_1-z_0)\overline{(z_2-z_0)}=R^2.$ From this, we can deduce that for any $z\ne 0$, the point symmetric to $z$ about the unit circle will be $\frac1{\overline z}.$ Also, $0$ is symmetric about the unit circle to the so-called "point at $\infty$." Thus, since $\zeta\mapsto 0,$ we need its symmetric point about the unit circle to map to $\infty.$ Note that we can't have $d=0$, for then $f(0)=\infty\notin\Delta,$ so in order to map $\zeta$'s symmetric point about the unit circle to $\infty,$ we need $c=-d\overline\zeta$. (Why?)
Let's use the values we've found for $b,c$ to rewrite $$f(z)=\frac{a(z-\zeta)}{d\left(\overline\zeta z-1\right)}=\frac ad\cdot\frac{z-\zeta}{\overline\zeta z-1}=A\frac{z-\zeta}{\overline\zeta z-1}.$$ Now, note that $$|f(z)|^2=|A|^2\frac{|z-\zeta|^2}{|\overline\zeta z-1|^2}=|A|^2\frac{|z|^2-2\Re(z\overline\zeta)+|\zeta|^2}{|\zeta|^2|z|^2-2\Re(z\overline\zeta)+1},$$ so when $|z|=1,$ we have $|f(z)|^2=|A|^2$. Since $f$ maps the unit circle to itself, then $|A|^2=1,$ so $|A|=1,$ and so $A=e^{i\alpha}$ for some real $\alpha$, and so $f$ has the form $(\clubsuit)$, as desired.
