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    I've been given the question:
    One root of the quadratic equation z^2 + az + b = 0 where a and b are real, is the complex number 1 + 2i.
    i) Write down the other root
    ii) Find the values of a and b
    I'm unsure how to answer this one.

    I also got the question:
    Find the quadratic equation which has roots 2 + 3i and 2 - 3i.
    For this I expanded (2+3i)(2-3i) and ended up with 13, which is obviously wrong, but I can't think of another way of doing it.

    Hopefully someone can help.
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    (Original post by CCauston113)
    I've been given the question:
    One root of the quadratic equation z^2 + az + b = 0 where a and b are real, is the complex number 1 + 2i.
    i) Write down the other root
    ii) Find the values of a and b
    I'm unsure how to answer this one.

    I also got the question:
    Find the quadratic equation which has roots 2 + 3i and 2 - 3i.
    For this I expanded (2+3i)(2-3i) and ended up with 13, which is obviously wrong, but I can't think of another way of doing it.

    Hopefully someone can help.
    First question: i) if 1+2i is a root, then 1-2i is also a root as complex numbers always come in conjugate pairs because of the quadratic formula ii) both 1+2i and 1-2i are a root, so z=1±2i. To find the equation: (z-(1+2i))(z-(1-2i)) = (z-1-2i)(z-1+2i) then expand the brackets, all the i will cancel out and you will be left with z^2 - 2z + 4 = 0, so compare this to z^2 + az + b = 0, and a = -2 and b = 4 Second question: do the same as above, so we know the roots are z=2±3i, so make brackets: (z-(2+3i))(z-(2-3i)) = (z-2-3i)(z-2+3i) and expand out so all the i cancel and you are left with a quadratic - which I have worked out but tell me what you get first and we can compare hope this helps
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    (Original post by megfield22)
    First question: i) if 1+2i is a root, then 1-2i is also a root as complex numbers always come in conjugate pairs because of the quadratic formula ii) both 1+2i and 1-2i are a root, so z=1±2i. To find the equation: (z-(1+2i))(z-(1-2i)) = (z-1-2i)(z-1+2i) then expand the brackets, all the i will cancel out and you will be left with z^2 - 2z + 4 = 0, so compare this to z^2 + az + b = 0, and a = -2 and b = 4 Second question: do the same as above, so we know the roots are z=2±3i, so make brackets: (z-(2+3i))(z-(2-3i)) = (z-2-3i)(z-2+3i) and expand out so all the i cancel and you are left with a quadratic - which I have worked out but tell me what you get first and we can compare hope this helps
    OK, thanks
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    (Original post by CCauston113)
    OK, thanks
    For the first one shouldn't it be z^2 - 2z + 5?

    For the second one I got z^2 - 4z + 13.
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    (Original post by CCauston113)
    For the first one shouldn't it be z^2 - 2z + 5?

    For the second one I got z^2 - 4z + 13.
    yes it is, thanks for spotting it was just a typo I think because I have +5 written down in my notes! thanks
 
 
 
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