# Proving on modulus and conjugates

1. Nov 15, 2005

### irony of truth

I am proving that (sqrt)2 |z| >= |Re z| + |Im z|. The professor gave this as an example, but I want to ask something...

Why is it that I can't use this strategy:
-> |Re z| <= |z|
-> |Im z| <= |z|
-> adding corresponding expression yields |Re z| + |Im z| <= 2|z|. (what's wrong here?)

2. Nov 15, 2005

### Tom Mattson

Staff Emeritus
You can't use it because $|Re z|+|Im z| \leq 2|z|$ simply does not imply that $|Re z| + |Im z| \leq \sqrt{2}|z|$. Simple example: Say $|Re z|+|Im z|=1.5$. The first inequality is satisfied, but the second is not.

3. Nov 15, 2005

### benorin

Try squaring both sides.

That is an upper bound for |Re z| + |Im z|, but not the least upper bound. Try squaring both sides.

4. Nov 15, 2005

### benorin

Oh yeah, put z=x+iy first.