# I this mathematically 'legal'?

1. Dec 3, 2011

### blahblah8724

When trying to bound this complex number using $|z| = R$, I turned the 'modulus of the square' into the 'square of the modulus', is this allowed?!

$|(z + i)^2|$

$= (|z+i|)^2$ 'modulus of the square' into the 'square of the modulus'

$\geq (|z| - |i| )^2$ triangle inequality variation

$= (|z| - 1)^2$

$= (R - 1)^2$

Help would be appreciated!

2. Dec 3, 2011

### dextercioby

Why don't you test it by direct computation ?

3. Dec 3, 2011

### spamiam

The key is to note that the function $f(x) = x^2$ is increasing on $(0, \infty)$ in the reals. Does that help?

4. Dec 3, 2011

### HallsofIvy

Because, while finding a counterexample would show it is NOT true, getting any number of examples where it does work would not prove it is true.

5. Dec 3, 2011

### I like Serena

Are you aware of the polar representation of complex numbers (aka Euler's formula)?
That is, $z=re^{i\phi}$, where $|z|=r$.

Perhaps you might consider what |z2| would be versus |z|2...?

6. Dec 3, 2011

### Citan Uzuki

Yes, blahblah8724, your reasoning is correct.

7. Dec 3, 2011

### disregardthat

Technically this is not allowed. The rule used here is seemingly a >= b --> a^2 > b^2 which is valid as a rule only if a,b > 0. In this case b= |z|-|i| could be negative. However, the triangle inequality also gives |z+i| >= |i|-|z|, so the conclusion is correct, but only after using both inequalities. Essentially: (a >= b and a >= -b) --> a^2 >= b^2.

8. Dec 3, 2011

### Dickfore

Good point. Although here it turns out you arrived at a correct inequality, as 'disregardthat' pointed out, this is due to:

$$\vert z_{1} - z_{2} \vert \ge \vert \vert z_1 \vert - \vert z_2 \vert \vert \ge 0$$

9. Dec 3, 2011

### Citan Uzuki

I was under the impression that that reasoning was already implicit, and was the reason for working with the squared terms in the first place.

10. Dec 3, 2011

### Dickfore

The squaring of the terms had nothing to do with it. The fact that we compare positive numbers (under the squaring operation) allows us to write the same inequality for the squares as well.

11. Dec 3, 2011

### Deveno

hmm...is it true that $|z|^2 = |z^2|$?

well, i'm notoriously weak-minded, so i will just see if this works:

let $z = a+ib$. then $|z|^2 = (\sqrt{a^2 + b^2})^2 = a^2 + b^2$.

now $|z^2| = |(a+ib)(a+ib)| = |(a^2 - b^2) + i(2ab)| = \sqrt{(a^2 - b^2)^2 + (2ab)^2}$

$$= \sqrt{a^4 - 2a^2b^2 + b^4 + 4a^2b^2} = \sqrt{a^4 + 2a^2b^2 + b^4} = \sqrt{(a^2+b^2)^2} = a^2 + b^2$$

huh. i guess so.

12. Dec 4, 2011

### Dickfore

It follows from a more general property of the modulus:

$$\vert z_1 \, z_2 \vert = \vert z_1 \vert \, \vert z_2 \vert$$