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FP1

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    hey guys, i have some difficulties when doing FP1, which says
    Find the square roots of 5-12i
    The following is my working
    Click image for larger version. 

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    finally i get the answer but i have one question
    i. why the final answer is not ±(3 +2i)?:confused:
    please someone help!
    by the way, the answer is ±(3 − 2i)
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    You're not far off the full solution, though you made a bit of a mess when solving the quartic.

    Firstly, a,b are real.

    So when you had your solutions for b, you can discard the two complex ones, and are just left with b=2 or -2.

    Then using a= -6/b, you can work out the "a" value in each case, and you then have the two desired solutions. Done.

    When solving the quartic, you should recognize that this is a quadratic in b^2. You could use a substitution of c=b^2, though it's not necessary.

    So factorising you get (b^2-4)(b^2+9) = 0

    Since b is real, b^2 is positive, and so the only solution is b^2=4, and hence b=+/- 2....

    Edit: In response to your last point.

    If b=2, then a=-3, so one solution is -3+2i
    If b=-2, then a=3, so other solution is 3-2i
    Hence +/- (3-2i)
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    You're too fast, ghostwalker ^^

    Anyways, another factoring tip you might like to know is that when you have a cubic with four terms, such as

    b^3 + 2b^2 + 9b + 18

    I always try to factor by grouping first. For example, if you realize that 1-to-2 is the same proportion as 9-to-18, then you don't have to do synthetic or long division. You can just identify the common factor of b+2 and factor it out.

    (b^3 + 2b^2) + (9b + 18)

    (b+2)(b^2) + (b+2)(9)

    Note: Another way to interpret this step is that you're taking out the common factors in each group and conveniently being left with two identical remains.

    (b+2)(b^2+9)
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    (Original post by aznkid66)
    You're too fast, ghostwalker ^^
    Wasn't quick enough - OP made a significant edit whilst I was working on a reply.
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    IIRC you always get 4 roots of a complex number for sone reason, so my answers would always be
    +/- a +/- bi.

    Never could get the hang of pulling out the +/-.


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by Snakefingers13)
    IIRC you always get 4 roots of a complex number for sone reason, so my answers would always be
    +/- a +/- bi.

    Never could get the hang of pulling out the +/-.


    This was posted from The Student Room's iPhone/iPad App
    Hm? When the value of 'a' is uniquely dependent (one-to-one correspondence) on the value of 'b', than the number of valid combinations is the number of solutions for 'b'.

    For example, if you had "I have 4 more cats than dogs, and I have ±2 dogs," then 2 dogs 6 cats would be a solution, as well as -2 dogs 2 cats. However, 2 dogs 2 cats and -2 dogs 6 cats are not solutions.

    As a rule, there are always two unique square roots (including complex) for any number, including complex numbers.
    For example, the unique roots of i=cis(pi/2) are cis(pi/4) and cis(5pi/4).
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    (Original post by Snakefingers13)
    IIRC you always get 4 roots of a complex number for sone reason
    Well, I'll have you know the Fundamental theorem of algebra disagrees.
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    Fair enough. My teacher sure screwed that one up.


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by aznkid66)
    You're too fast, ghostwalker ^^

    Anyways, another factoring tip you might like to know is that when you have a cubic with four terms, such as

    b^3 + 2b^2 + 9b + 18

    I always try to factor by grouping first. For example, if you realize that 1-to-2 is the same proportion as 9-to-18, then you don't have to do synthetic or long division. You can just identify the common factor of b+2 and factor it out.

    (b^3 + 2b^2) + (9b + 18)

    (b+2)(b^2) + (b+2)(9)

    Note: Another way to interpret this step is that you're taking out the common factors in each group and conveniently being left with two identical remains.

    (b+2)(b^2+9)
    Nice post.

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