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Math Help - please help me

  1. #1
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    please help me

    use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
    x{^4} + 1
    ---------------------------------------------------------------------------------------------------------------
    sketch :
     \mid Arg z \mid \leq \frac{\pi}{4} and determine if the set is :
    open,closed,bounded,connected ,simply connected , domain,compact or region
    answer :
    the given set is :
    closed,connected ,simply connected ,and region
    but not :
    open,bounded,domain,and compact
    my question : if the question : arg z not Arg z"principle argument" the answer is the same or not ???
    Last edited by flower3; July 10th 2009 at 11:30 AM.
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  2. #2
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    From the law a^2-b^2=(a+b)(a-b), the fact that i^2=-1, and the fact that x^4+1=x^4-(-1), you should be able to prove the first part easily.
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  3. #3
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    Hello, flower3!

    Use complex analysis to factor the following polynomial
    into a product of two quadratic polynomials: . x{^4} + 1
    Complex analysis? . . . It requires only Algebra 1.


    Add and subtract 2x^2\!:\quad x^4 \:{\color{red}+ \:2x^2} + 1 \:{\color{red}- \:2x^2} \;=\;\left(x^2+1\right)^2 - \left(\sqrt{2}\:\!x\right)^2


    We have a difference of squares:

    . . \left[(x^2+1) - \sqrt{2}\:\!x\right]\,\left[(x^2+1) + \sqrt{2}\:\!x\right]

    . . . =\;\left(x^2 - \sqrt{2}\:\!x + 1\right)\,\left(x^2 + \sqrt{2}\:\!x + 1\right)

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  4. #4
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    Quote Originally Posted by flower3 View Post
    use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
    x{^4} + 1
    x^4 + 1

    = x^4-i^2

    = (x^2+i)(x^2-i)
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  5. #5
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    Quote Originally Posted by malaygoel View Post
    x^4 + 1

    = x^4-i^2

    = (x^2+i)(x^2-i)
    I suspect the OP meant two quadratic polynomials with real coefficients.

    Quote Originally Posted by Soroban View Post
    Hello, flower3!

    Complex analysis? . . . It requires only Algebra 1.


    Add and subtract 2x^2\!:\quad x^4 \:{\color{red}+ \:2x^2} + 1 \:{\color{red}- \:2x^2} \;=\;\left(x^2+1\right)^2 - \left(\sqrt{2}\:\!x\right)^2


    We have a difference of squares:

    . . \left[(x^2+1) - \sqrt{2}\:\!x\right]\,\left[(x^2+1) + \sqrt{2}\:\!x\right]

    . . . =\;\left(x^2 - \sqrt{2}\:\!x + 1\right)\,\left(x^2 + \sqrt{2}\:\!x + 1\right)
    Unfortunately, in an exam setting, if the question gives the instruction to use complex analysis then a correct answer using an alternative technique will not score any marks. Not even an answer mark.

    Quote Originally Posted by flower3 View Post
    use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
    x{^4} + 1
    [snip]
    Start by finding the four complex roots of -1. Can you do this? (If not, see http://www.mathhelpforum.com/math-he...ex-number.html, for example).
    Last edited by mr fantastic; July 12th 2009 at 05:12 AM.
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  6. #6
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    Quote Originally Posted by flower3 View Post
    [snip]
    sketch :
     \mid Arg z \mid \leq \frac{\pi}{4} and determine if the set is :
    open,closed,bounded,connected ,simply connected , domain,compact or region
    answer :
    the given set is :
    closed,connected ,simply connected ,and region
    but not :
    open,bounded,domain,and compact
    my question : if the question : arg z not Arg z"principle argument" the answer is the same or not ???
    Your question can be easily answered by noting that  \mid \text{Arg} (z) \mid \leq \frac{\pi}{4} is equivalent to  - \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4} ....
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Your question can be easily answered by noting that  \mid \text{Arg} (z) \mid \leq \frac{\pi}{4} is equivalent to  - \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4} ....
    i know that , but i mean if the question  \mid arg z  \mid \leq \frac{\pi}{4} is the answer the same or not ???
    closed,connected ,simply connected ,and region
    but not :
    open,bounded,domain,and compact
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    I suspect the OP meant two quadratic polynomials with real coefficients.


    Unfortunately, in an exam setting, if the question gives the instruction to use complex analysis then a correct answer using an alternative technique will not score any marks. Not even an answer mark.


    Start by finding the four complex roots of -1. Can you do this? (If not, see http://www.mathhelpforum.com/math-he...ex-number.html, for example).
    Thanks !
    if i find one of theses roots and then multiply it by its conjugate then by using long division i find the 2nd quadratic polynomial "true or not ???"
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  9. #9
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    Quote Originally Posted by flower3 View Post
    Thanks !
    if i find one of theses roots and then multiply it by its conjugate then by using long division i find the 2nd quadratic polynomial "true or not ???"
    The roots will have the form z = \alpha_1, z = \overline{\alpha_1}, z = \alpha_2 and z = \overline{\alpha_2}.

    Therefore the two quadratic factors you require will obviously be (z - \alpha_1)(z - \overline{\alpha_1}) and (z - \alpha_2)(z - \overline{\alpha_2}).

    Quote Originally Posted by flower3 View Post
    i know that , but i mean if the question  \mid arg z \mid \leq \frac{\pi}{4} is the answer the same or not ???
    closed,connected ,simply connected ,and region
    but not :
    open,bounded,domain,and compact
    The answer is contained in post #6.
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