# Thread: Prove that the modulus is a norm on C

1. ## Prove that the modulus is a norm on C

Folks,

I have $z=a + ib, |z|=\sqrt{a^2+b^2}$

Axiom 1: $|z| \ge0$

Axiom 2: $|z|=0$ iff $a=b=0$

Axiom 3: $| \alpha| |z|=|\alpha z|$ for $\alpha \in \mathbb{C}$

Axiom 4: The triangle inequality.

How do I get started on any of these axioms?

THanks

2. ## Re: Prove that the modulus is a norm on C

is |z| ≥ 0, for any complex number z? certainly $a^2+b^2 \geq 0$ for any real a,b (why?). thus the positive square root is defined for such a number. why is this non-negative? these are just basic properties of real numbers.

when can |z| be 0? draw a picture. formalize your picture with a proof.

for axiom 3, write out α and z as α = c+id, z = a+ib. do the multiplication. take the norm of the product, and compare with the product of the norms. what do you find?

you have to use your definitions to prove axioms.

3. ## Re: Prove that the modulus is a norm on C

Originally Posted by Deveno
is |z| ≥ 0, for any complex number z? certainly $a^2+b^2 \geq 0$ for any real a,b (why?). thus the positive square root is defined for such a number. why is this non-negative?
I dont know how to show |z| \ge 0 despite what you say about a and b....?

Originally Posted by Deveno
when can |z| be 0? draw a picture. formalize your picture with a proof.
|z| can be 0 only when the function z is 0, ie is the 0 function...?

Originally Posted by Deveno
for axiom 3, write out α and z as α = c+id, z = a+ib. do the multiplication. take the norm of the product, and compare with the product of the norms. what do you find?
I get

$| a z |=|(ac-bd)+(ad+bc)i|=\sqrt{(ac-bd)^2+((ad+bc)i)^2}$ and

$|a||z|=|c+id|+|a+ib|=\sqrt{c^2+(id)^2}\sqrt{a^2+(i b)^2}$.......?

4. ## Re: Prove that the modulus is a norm on C

Originally Posted by bugatti79
I dont know how to show |z| ≥ 0 despite what you say about a and b....?
um....isn't the POSITIVE square root of a non-negative number always 0 or positive?

|z| can be 0 only when the function z is 0, ie is the 0 function...?
you're just guessing here. when is √x = 0? how many real numbers have this property? if (aČ + bČ) = 0, what could a and b possibly be? could a be -4? 6? 117? how about b?

z is not a "function". it's just a plain ol' complex number.

I get

$| a z |=|(ac-bd)+(ad+bc)i|=\sqrt{(ac-bd)^2+((ad+bc)i)^2}$ and

$|a||z|=|c+id|+|a+ib|=\sqrt{c^2+(id)^2}\sqrt{a^2+(i b)^2}$.......?
there's no "i" terms in the definition of modulus, look again.

5. ## Re: Prove that the modulus is a norm on C

In order to prove the triangle inequality property of the complex norm you must prove some lemmas.
$1)~|z|^2=z\cdot\overline{z}$.

$2)~|\text{Re}(z)|\le |z|$

$3)~z\cdot\overline{w}+\overline{z}\cdot w=2\text{Re}(z\cdot\overline{w})$.

Once you have shown those are true then consider:
$|z+w|^2$.

6. ## Re: Prove that the modulus is a norm on C

Originally Posted by bugatti79
Folks,

I have $z=a + ib, |z|=\sqrt{a^2+b^2}$

Axiom 4: Prove the triangle inequality.

THanks
Folks, for this axiom the notes starts of proving that

$|z_1+z_2| \le |z_1|+|z_2|$ (1) where $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

after some algebra, it arrives at

$x_1x_2+y_1y_2 \le|z_1||z_2|$ and it mentions the Cauchy Schwarz Ineq but how does it prove eqn (1)?