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Math Help - Prove that the modulus is a norm on C

  1. #1
    Senior Member bugatti79's Avatar
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    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
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  2. #2
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    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.
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  3. #3
    Senior Member bugatti79's Avatar
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    Re: Prove that the modulus is a norm on C

    Quote Originally Posted by Deveno View Post
    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 View Post
    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 View Post
    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}.......?
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  4. #4
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    Re: Prove that the modulus is a norm on C

    Quote Originally Posted by bugatti79 View Post
    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.
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  5. #5
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    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.
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  6. #6
    Senior Member bugatti79's Avatar
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    Re: Prove that the modulus is a norm on C

    Quote Originally Posted by bugatti79 View Post
    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)?
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