Proving on modulus and conjugates

[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?)

 

[quantumdude]

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?)

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.

 

[benorin]

Try squaring both sides.

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?)

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

 

[benorin]

Oh yeah, put z=x+iy first.