# Convexity of Open Disc

1. Feb 25, 2008

### e(ho0n3

[SOLVED] Convexity of Open Disc

I'm trying to prove that the open disc in the complex plane given by D = {z : |z - w| < r} is convex.

Let p and q be two points in D. The line segment from p to q is L = {(1 - t)p + tq : 0 <= t <= 1}. Let u be a point an arbitrary point on this segment. If I can show that |u - w| < r, I'm done.

This is essentially a geometry problem. There's probably a proposition in Euclid's Elements that the line segment from any two points on the circle is contained in the circle. How do you show this analytically though?

2. Feb 25, 2008

### Dick

Use the triangle inequality on w(t)=|(1-t)p+tq|. The result is a linear function in t. |w(0)|=|p|<=1 and |w(1)|=|q|<=1. Hence?

3. Feb 26, 2008

### e(ho0n3

Using the triangle inequality is a great idea:

|p| - |w| < |p - w| < r so |p| < r + |w|. Similarly |q| < r + |w|.

|(1 - t)p + tq - w| <= (1 - t)|p| + t|q| + |w| < (1 - t)(r + |w|) + t(r + |w|) + |w| = r + 2|w| so (1 - t)|p| + t|q| - |w| < r.

However, since (1 - t)|p| + t|q| - |w| <= |(1 - t)p + tq - w|, I don't know if the latter is greater than r.

4. Feb 26, 2008

### Dick

I can how my first reply could be a little difficult to digest. I was taking w=0 and r=1 and not telling you that. Furthermore I meant to define u(t)=p(1-t)+qt to fit in with your notation better. Let's try that again. u(t)-w=(p-w)(1-t)+(q-w)t, right? So |u(t)-w|<=|p-w|(1-t)+|q-w|t. Can you finish it from there?

5. Feb 26, 2008

### e(ho0n3

Ah, I see. Writing u(t)-w as (p-w)(1-t)+(q-w)t is the right idea. It never occurred to me. From then on, it trivially follows that |u - w| < r. Thanks a lot.