A problem on modulus of complex numbers Let $ z,a \in \Bbb C $ and suppose that $ |a| <1$ Prove that $ |z-a| \leq |1- {\bar a} z| $ iff $|z| \leq 1 $
I couldn't find a way to approach the sum without using $a,z$ and $x+iy$, from which the answer can be obtained but it is tedious, wondered if there is any way to derive from the properties of modulus and conjugate
Any help would be appreciated
Thank You
 A: Just expand the given inequality and simplify:
$$|z-a| \leq |1 - \overline{a}z|$$
iff
$$|z-a|^2 \leq |1 - \overline{a}z|^2$$
iff
$$|z|^2 - \overline{z}a - \overline{a}z + |a|^2 \leq 1 - \overline{a}z - \overline{z}a + |a|^2 |z|^2$$
iff
$$|z|^2 + |a|^2 \leq 1 + |a|^2|z|^2$$
iff
$$0 \leq 1 - |z|^2 - |a|^2 + |a|^2|z|^2$$
iff
$$0 \leq (1 - |z|^2)(1 - |a|^2)$$
You are given that $|a| < 1$, so the result follows.
A: The inequality says:
$$
\left|\frac{z-a}{1-\bar{a} z}\right| \le 1 \text{ if and only if }|z| \le 1.
$$
So that means the mapping
$$
z\mapsto \frac{z-a}{1-\bar{a} z} \tag 1
$$
is a bijection from the disk $|z|\le1$ to itself.  This mapping is a continuous bijection from the Riemann sphere $\mathbb C\cup\{\infty\}$ to itself. If its restriction to the circle $|z|=1$ is a bijection to the circle $|z|=1$, then it must map the disk $|z|\le 1$ either to itself or to the complementary disk $|z|\ge1$.  Since the mapping $(1)$ takes $0$ to $a$, and $a$ is inside the disk $|z|=1$, the mapping $(1)$ must therefore take the disk $|z|\le1$ bijectively to itself provided only that we show that it takes the circle $|z|=1$ bijectively to itself.
Linear fractional transformations take circles in the Riemann sphere to circles in the Riemann sphere.  So now all we have to do is find out which circle the unit circle $|z|=1$ is mapped to.  To do that, it's enough to find the images of three points.  Notice that the two points $z$ satisfying $z^2=a/\bar a$ are on the circle and are fixed points of $(1)$.  That gets you two points; now find a third.
A: The solution is based on the fact that for any two real numbers $r$ ans $s$ the product $(1-r)(s-1)=r+s-rs-1$ is $\leq 0$ only if both $r,s \leq 1$ or both $r,s\geq 1$.
Now we have $|z-a| \leq |1-\bar{a}z| \Leftrightarrow |z-a|^2 \leq |1-\bar{a}z|^2\Leftrightarrow z\bar{z}+ a\bar{a}-(a\bar{z}+a\bar{z})\leq 1+a\bar{z}a\bar{z}-(a\bar{z}+a\bar{z})\Leftrightarrow z\bar{z}+a\bar{a}-a\bar{a}z\bar{z}-1\leq 0$. If we substitute $r=z\bar{z}$ and $s=a\bar{a}$ then the initial remark proves the assertion. And even more: if $a\bar{a}\geq1$ then als $z\bar{z}\geq 1$ 
