Check that this transformation is a contraction Let $T_0, T_1:\mathbb{C}\to\mathbb{C}$ be given by $T_0(z) = rz\exp{\frac{i\pi}{4}}+i$ and $T_1(z) = rz\exp{\frac{-i\pi}{4}}+i$ respectively, for $0<r<1$. I need to show that these are contractions. I'm taking this to mean that $|T_0(z)|<|z|$ and $|T_1(z)| < |z|$, $\forall z \in \mathbb{C}$

Note: This is in the context of binary trees where each branch of the tree leads to a transformation of the previous branching point. For example, the first point might be just $i$, and the next two points on the tree are $T_0(i)$ and $T_1(i)$. Then the process is applied to each of those points, and to the resulting points, and so on.

Let's first look at $T_0$.
Using Euler's formula we can say $$\exp{\frac{i\pi}{4}} = \cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$$
So we have $$T_0(z) = rz\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)+i$$
If we let $z = a+bi$, then $|z| = \sqrt{a^2 + b^2}$. Subbing $a+bi$ in for $z$ in $T_0(z)$, we get $$T_0(z) = r\left(a+bi\right)\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)+i$$
$$=\frac{r\sqrt{2}}{2}\left(a-b+(a+b)i\right)+i$$
$$=\frac{r\sqrt{2}}{2}\left(a-b\right)+\left(\frac{r\sqrt{2}}{2}(a+b)+1\right)i$$
Then $$|T_0(z)| = \sqrt{\left(\frac{r\sqrt{2}}{2}(a-b)\right)^2+\left(\frac{r\sqrt{2}}{2}(a+b)+1\right)^2}.$$
Similarly
$$|T_1(z)| = \sqrt{\left(\frac{r\sqrt{2}}{2}(a-b)\right)^2+\left(\frac{r\sqrt{2}}{2}(b-a)+1\right)^2}.$$
I don't think I can prove that these are less than $|z|$ for all $z$.
Am I missing the correct interpretation of the question?
 A: We can address both $T_0$ and $T_1$ together by defining
$T_\pm(z) = rze^{\pm i\pi/4} + i; \tag 1$
then
$\vert T_\pm(z_1) - T_\pm(z_2) \vert = \vert rz_1 e^{\pm i\pi/4} + i  - rz_2 e^{\pm i\pi/4} - i \vert$
$= \vert r(z_1 - z_2) e^{\pm i\pi/4} \vert; \tag 2$
now,
$\vert r(z_1 - z_2) e^{\pm i\pi/4} \vert = \vert r \vert \vert z_1 - z_2 \vert \vert e^{\pm i \pi/4} \vert, \tag 3$
since $\vert \cdot \vert$ is multiplicative, that is,
$\vert ab \vert = \vert a \vert \vert b \vert, \; \forall a, b \in \Bbb C; \tag 4$
furthermore, $0 < r \in \Bbb R$ and so
$\vert r \vert = r, \tag 5$
and
$\vert e^{\pm i\pi/4} \vert = 1; \tag 6$
thus (3) becomes
$\vert r(z_1 - z_2) e^{\pm i\pi/4} \vert = \vert r \vert \vert z_1 - z_2 \vert \vert e^{\pm i \pi/4} \vert$
$= r\vert z_1 - z_2 \vert < \vert z_1 - z_2 \vert; \tag 7$
which shows that each of $T_\pm$, that is, $T_0$ and $T_1$, are contractions.
Nota Bene:  Why do we have
$\vert e^{\pm i \pi /4} \vert = 1? \tag 8$
well,
$e^{\pm i \pi /4} = \cos (\pi /4) \pm i \sin(\pi /4), \tag 9$
and
$\cos^2 (\pi /4) + \sin^2(\pi /4) = 1, \tag{10}$
that is,
$\vert e^{\pm i \pi /4} \vert^2 = 1, \tag{11}$
from which (8) immediately follows.  End of Note.
