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    lxl < 2 <=> -2 < x < 2 ??? Correct - yes/no ???
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    yep
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    if x is real then yes
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    (Original post by mik1w)
    if x is real then yes
    Well "<" and ">" has no meaning in complex numbers, since you cannot order the complex plane, so it has to be real x.
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    (Original post by AlphaNumeric)
    Well "<" and ">" has no meaning in complex numbers, since you cannot order the complex plane, so it has to be real x.
    I dont know about this, but surely if you take the modulus of a complex number, you end up with a real number, meaning that you can test whether the modulus is less than 2?
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    Yes, if you take the modulus of a complex number, then you are effectively mapping it to the real numbers, which are ordered, so you can apply the > or < equivalence relation.

    However you cannot apply the < or > signs to complex numbers directly. After all, is "i" greater than or less than 2? The modulus of i is certainly less than the modulus of 2, but if you don't take the modulus you cannot say which is greater than the other. This is because you cannot order the complex numbers.
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    well, I am abit confused about it, not about comparison, but about the modulus.
    In Alevel, its notation is |z|, but in Uni, they call it norm of z and write ||z||, or is it only in my Uni?
    If ||z|| is right, then there's no meaning for |z| ?
    Can you clear it out for me?
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    Doesnt that mean the initial assertion should have been

    |x| => -2 < x < 2

    if no other information(such as x ER) is available
    because if x is complex, that works, whereas and if and only if relationship wont work, as because you pointed out, the greater than or less than operators do not apply for complex x.
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    Possibly the double mod is used because since the definition of a complex modulus gives a square root, which can be positive or neg, it gives us information to take the +ve root.
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    It is much of a muchness. ||z|| is a more general term, because there happen to be infinitely many norms

    If z = (z_{1},z_{2},...,z_{N}) then ||z||_{n} \equiv \Big[ \sum_{i=1}^{N} z_{i}^{n} \Big]^{\frac{1}{N}}

    ||z||_{1} = z_{1}+z_{2}+...z_{N}
    ||z||_{2} = \sqrt{z_{1}^{2}+z_{2}^{2}+...+z_  {N}^{2}} = |z|

    ||z||_{2} is the "usual" norm (its the pythagoras formula if you write it out), and |z| is the short hand noration for this. Hence, if you write |z| then you mean ||z||_{2}. If you write ||z|| then it is possible you might be referring to a different, or more general norm. Typically this only happens if you're doing questions about norms though, in say an analysis course. If I saw someone write ||z|| in a methods question I'd just take it to mean |z| unless they'd defined it to be something else.
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    (Original post by Hitonagashi)
    Possibly the double mod is used because since the definition of a complex modulus gives a square root, which can be positive or neg, it gives us information to take the +ve root.
    What? The modulus of complex can be -ve ?
    Isn't it a length of vector representing the complex number?
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    Exactly. However, a square root can obviously give +ve and -ve roots, so my theory was the double mod prevented it from being negative, but AN gave a much better explanation
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    (Original post by BCHL85)
    What? The modulus of complex can be -ve ?
    Isn't it a length of vector representing the complex number?
    No, the mod of a number can NEVER be negative, irrelevant of the use of |z| or ||z|| notation.

    Norms obey the following rules :

    ||z|| &gt; 0  \quad \forall z
    ||z|| + ||x|| \geq ||x+z|| (otherwise known as "triangle inequality")
    ||z-x|| = 0 &lt;=&gt; x=z
    ||x-z|| = ||z-x||

    If you're interested, check out Chapter 2 here. There's a tiny bit about ordered fields right at the beginning of Chapter 1 too
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    ooh...interesting. Thanks for the link.
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    Thanks AlphaNumeric, but in your notation z = (z1, z2 ,.... , zN). What are z1, z2,...?
    Are they real or complex. If complex then I can't understand how you get ||z||n

    --------------

    thanks for the link as well
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    It's vector notation. In 3D you'd have a vector z = (z_{1},z_{2},z_{3}). If z is complex, z=x+iy then you could denote z as z = (x,y)
 
 
 

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