# Points in the unit disk

Suppose we have complex numbers $$\alpha_1,..., \alpha_n$$ that lie on the unit disk. Let $$t_1,...,t_n$$ be positive numbers such that $$t_1 + ... + t_n =1$$. Then $$\alpha_1 t_1 + ... + \alpha_n t_n$$ lies in the unit disk.

### Try

We have by hypothesis $$|\alpha_i | < 1$$ for all $$i = 1,...,n$$. I was thinking about using Cauchy Schwarz inequality:

$$|\alpha_1 t_1 + ... + \alpha_2 t_2 |^2 \leq ( |\alpha_1|^2 + ... | \alpha_n|^2)( |t_1|^2 + ... |t_n|^2 )$$

But it seems this does not give me a good estimate. Is there another approach to solve this problem ?

## 5 Answers

stop thinking complex and start thinking planar geometry.

if you take two points inside the unit circle. The set of points of the form you give is the straight line between those points. So you are saying is that the unit disc is convex.

The general case follows by induction.

Using the triangle inequality and $|\alpha_i|<1$,

$$|\alpha_1 t_1 + \cdots + \alpha_n t_n| \leq |\alpha_1| t_1 + \cdots |\alpha_n |t_n < t_1 + \cdots t_n = 1$$

By induction on $n$. I denote the unit disk by $D$.

If $n=1$, this is trivial.

Suppose this holds for $n-1$ points. Now $$\frac{t_1}{1-t_n} + \dots \frac{t_{n-1}}{1-t_n} = 1$$

so by inductive hypothesis

$$\frac{t_1}{1-t_n} \alpha_1+ \dots \frac{t_{n-1}}{1-t_n}\alpha_{n-1} \in D$$

and finally, since $D$ is convex

$$(1-t_n) \left( \frac{t_1}{1-t_n} \alpha_1+ \dots \frac{t_{n-1}}{1-t_n}\alpha_{n-1} \right) + t_n \alpha_n\in D$$ and this completes the proof.

Hint:

Try induction over $n$:

For $n = 1$ you have $t_1\alpha_1 = 1$ where $t_1 = 1$ and $\alpha_1$ lies on the unit disc.

Try proving it for $n=2$.

Then if it works for $n-1$ then let $\beta = t_1\alpha_1 + t_2\alpha_2 + \ldots + t_{n-1}\alpha_{n-1}$ and $u = t_1 + \ldots t_{n-1}$ then:

$u + t_n = 1$ and $\beta$ and $\alpha_n$ lie on the unit disc. (Using the induction hypothesis for $\beta$). This is essentially the same case as in $n=2$.

Try using induction on the number of points. Trivially true with one point and easily true with two (because the weighted sum is a point on the line segment joining the points, and the unit disk is convex). For $n > 2$, let $v = 1/(1-t_n)$. Then $$a_1t_1 + \cdots + a_nt_n = (a_1t_1v + \cdots + a_{n-1}t_{n-1}v)(1-t_n) + a_nt_n$$ By induction, the first term is a point on the disk, reducing the $n$-point problem to a 2-point problem, which you've already done.