Prove that the modulus is a norm on C

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November 10th 2011, 10:27 AM
bugatti79

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
November 10th 2011, 10:33 AM
Deveno

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.
November 10th 2011, 11:12 AM
bugatti79

Re: Prove that the modulus is a norm on C

Quote:

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....?

Quote:

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...?

Quote:

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

$|a||z|=|c+id|+|a+ib|=\sqrt{c^2+(id)^2}\sqrt{a^2+(i b)^2}$ .......?
November 10th 2011, 12:06 PM
Deveno

Re: Prove that the modulus is a norm on C

Quote:

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?

Quote:

|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 (a² + b²) = 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.

Quote:

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.
November 10th 2011, 12:42 PM
Plato

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:
$|z+w|^2$ .
November 17th 2011, 11:22 AM
bugatti79

Re: Prove that the modulus is a norm on C

Quote:

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)?