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Every complex number is the square of another... watch

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    I'm pretty sure this is true but I thought I'd check just in case:

    'Every complex number is the square of another complex number'.

    It seems true because every number x+iy can be written as (u+iv)^2 where x=u^2-v^2 and y=2uv.
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    There is closure, of course assuming that you count the reals as within the complex set, yup.
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    (Original post by 0-))
    I'm pretty sure this is true but I thought I'd check just in case:

    'Every complex number is the square of another complex number'.

    It seems true because every number x+iy can be written as (u+iv)^2 where x=u^2-v^2 and y=2uv.
    It is true. Have you encountered polar form yet?
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    (Original post by generalebriety)
    It is true. Have you encountered polar form yet?
    Yup.
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    That's pretty cool never thought of that before!

    (By the way, is it not slightly sad that this arose at past 1am on a monday morning??)lol!
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    (Original post by qwerty5102)
    That's pretty cool never thought of that before!

    (By the way, is it not slightly sad that this arose at past 1am on a monday morning??)lol!
    I was trying to get to sleep when this thought occured to me and I had to enquire.

    I dream about maths! :cool:
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    (Original post by 0-))
    Yup.
    Cool. Well, if you write your number as r(cos A + i sin A), then one of its square roots is sqrt(r) (cos A/2 + i sin A/2) (remember that squaring a number squares its modulus and doubles its argument).
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    (Original post by generalebriety)
    Cool. Well, if you write your number as r(cos A + i sin A), then one of its square roots is sqrt(r) (cos A/2 + i sin A/2) (remember that squaring a number squares its modulus and doubles its argument).
    Ah yes - cheers for that.

    I can now sleep peacefully .
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    One last question:

    Do all complex poynomials of degree n have n complex roots?

    It's been true for all polys I've encountered so I predict it's true.
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    (Original post by 0-))
    One last question:

    Do all complex poynomials of degree n have n complex roots?

    It's been true for all polys I've encountered so I predict it's true.
    Yes, though they're not necessarily distinct. (x-3)^5, for example, has five roots: 3, 3, 3, 3 and 3. This is called the fundamental theorem of algebra.
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    (Original post by generalebriety)
    Yes, though they're not necessarily distinct. (x-3)^5, for example, has five roots: 3, 3, 3, 3 and 3. This is called the fundamental theorem of algebra.
    Thanks again.

    One more question (I promise ):

    Are inequalities ever used as relations between complex numbers?
    e.g. 2i > i

    And if they are used, how is each number defined in relation to another?
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    (Original post by 0-))
    Thanks again.

    One more question (I promise ):

    Are inequalities ever used as relations between complex numbers?
    e.g. 2i > i

    And if they are used, how is each number defined in relation to another?
    Yes, but we define the equations in terms of an argument and the modulus on an argand diagram. Look up loci on an argand diagram.
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    (Original post by 0-))
    Thanks again.

    One more question (I promise ):

    Are inequalities ever used as relations between complex numbers?
    e.g. 2i > i

    And if they are used, how is each number defined in relation to another?
    No, absolutely not. As HMK says, you can compare the sizes of moduli and arguments (because they're just real numbers), but there's no such thing as an order on the complex plane itself. (From a philosophical point of view, it's to do with our inability to distinguish between -i and +i. Let i^2 = -1, and let j = -i. Then we have two roots of -1, namely i and j. Both are negatives of each other, both give -1 when squared... in fact, they're completely indistinguishable mathematically. Compare this with -1 and 1, in some sense both "square roots" of 1 - we can distinguish these because 1^2 is itself and (-1)^2 isn't itself.)
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    (Original post by generalebriety)
    No, absolutely not. As HMK says, you can compare the sizes of moduli and arguments (because they're just real numbers), but there's no such thing as an order on the complex plane itself. (From a philosophical point of view, it's to do with our inability to distinguish between -i and +i. Let i^2 = -1, and let j = -i. Then we have two roots of -1, namely i and j. Both are negatives of each other, both give -1 when squared... in fact, they're completely indistinguishable mathematically. Compare this with -1 and 1, in some sense both "square roots" of 1 - we can distinguish these because 1^2 is itself and (-1)^2 isn't itself.)
    Thank you - some interesting info there.

    Also, thank you HMK.
 
 
 
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