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    Can anyone give any tips/techniques when factorising complext equations
    e.g:
    z^2+6z+13,
    z^2+z+1,
    z^3+1

    Many Thanks
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    Use the quadratic formula for the first 2, and for the 3rd, take out a factor of (z-1) first then it leaves another quadratic.
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    Thanks :rolleyes:
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    For the 3rd, i meant (z+1) as the factor
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    Here's a general method for solving equations like the 3rd one.

    f(x) = x³+ax²+bx+c
    Take the factors of c and use them in f(x). If the result is zero, that number is a factor.
    eg:
    f(x) = x³+1
    Factors of one: 1 (and -1)
    f(1) = 1+1 = 2, so this isn't a factor.
    f(-1) = -1+1 = 0, so this is a factor.

    So.. x=-1 => (x+1)=0
    Now divide x³+1 by x+1 to get a quadratic equation, and factorize that to get all the required factors.
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    Since you're dealing with complex numbers, remember you can use the following identity,

    z² + a² = (z + ia)(z - ia)
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    (Original post by Fermat)
    Since you're dealing with complex numbers, remember you can use the following identity,

    z² + a² = (z + ia)(z - ia)
    Complex equations.
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    (Original post by sweetymango)
    Can anyone give any tips/techniques when factorising complext equations
    e.g:
    z^2+6z+13,
    z^2+z+1,
    z^3+1

    Many Thanks
    z = a + bj

    z^2+6z+13
    = (a+bj)^2 + 6(a + bj) + 13
    = a^2 + 2abj - b^2 + 6a + 6bj + 13
    = (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

    Is that how you mean?
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    I don't think s/he meant the mathematical definition of "complex".
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    (Original post by shift3)
    Complex equations.
    Well, complex equations will involve complex numbers, and should you have the complex equation,

    z² + a² = w²

    then this can be factorised and rewritten as,

    (z+ia)(z-ia) = w²
    ==========
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    I think the person means complex as in difficult and not the mathematical sense Fermat...
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    (Original post by ZJuwelH)
    I think the person means complex as in difficult and not the mathematical sense Fermat...
    Aaaargh!
    I think I'll go back to sleep.
    Silly Zzzzzzzz....
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    I've only ever seen z notation in the P4 complex numbers chapter, so surely Mr. Fermat was correct in his efforts?
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    Of course they are complex functions, because the variable z is used.
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    Can anyone give any tips/techniques when factorising complext equations
    e.g:
    z^2+6z+13,
    z^2+z+1,
    z^3+1
    The above are functions of z, not equations. Accepting the standard notation that z = x+iy where x,y are real numbers and i = sqrt(-1) then they can, correctly, be described as complex functions.
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    to clear things up, i am using z as a complex number i.e in the form of x+yj
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    (Original post by benm)
    z = a + bj

    z^2+6z+13
    = (a+bj)^2 + 6(a + bj) + 13
    = a^2 + 2abj - b^2 + 6a + 6bj + 13
    = (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

    Is that how you mean?
    well i meant writing them in the form as products of its factors, I've found that the method is to use the quadratic formula
 
 
 
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