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    Obtain a Quadratic Function f(z) = z^2 + az + b , where a,b member of R such that f(-1-2i) = 0

    I know that the conugate is f(-1+2i) , but I just dont know what to do from then on? All help appreciated. Thanks
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    (Original post by xyz2k6)
    Obtain a Quadratic Function f(z) = z^2 + az + b , where a,b member of R such that f(-1-2i) = 0

    I know that the conugate is f(-1+2i) , but I just dont know what to do from then on? All help appreciated. Thanks
    You requre:
    (-1-2i)^2+a(-1-2i)+b=0
    Multiply out, factorise by the real and imaginary components, then solve for a and b.
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    z = -1+2i and z = -1-2i are the roots

    so, z-(-1+2i) and z-(-1-2i) are the factors
    ie. (z+1-2i) and (z+1+2i) are the factors

    (z+1-2i)(z+1+2i) = z2 + 2z + 5
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    This has kind of already been said, but two very important things to remember are

    (1) Complex roots come in conjugate pairs. If a+bi (b nonzero) is a root, so is a-bi. This is because for a real polynomial p, p(a-bi) is the conjugate of p(a+bi).
    (2) If f(a)=0 then (x-a) divides f(x).

    Can you tell me why a real cubic cannot have precisely two real roots?
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    (Original post by mikesgt2)
    Can you tell me why a real cubic cannot have precisely two real roots?
    Think about the shape of a real cubic. It can never intercept the x-axis exactly twice. Possibilities are:

    1) It crosses the x-axis once, turns above the x-axis, turns again above the x-axis, and then goes off to infinity.
    2) It crosses the x-axis once, turns again and crosses the x-axis again, and then it must turn AGAIN and cross the x-axis again.

    The closest you can get to having two real roots is if it turns a second time and the turning point is ON the x-axis. In this case you will get one root where it crosses the first time, and a repeated root where it just touches.

    I'm sure you can prove this somehow, but I'm a physicist - it seems obvious to me, but if you want algebraic proof, I'm not going to do it!
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    Worzo, of course what you say is correct. However, if you wanted to prove it without appealing to what the graph looks like you could use number (1) in my post. What can you say about the number of complex roots a polynomial has?
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    (Original post by mikesgt2)
    Worzo, of course what you say is correct. However, if you wanted to prove it without appealing to what the graph looks like you could use number (1) in my post. What can you say about the number of complex roots a polynomial has?
    A cubic is of degree 3, therefore, by the fundamental theorem of algebra, has 3 roots. If it were to have 2 real roots, then it would have to have (3-2=1) one complex root. But by (1), if it has one complex root, it must have another - its complex conjugate. This is contradiction. Therefore, a cubic cannot have 2 real roots.

    Were you actually asking the question or trying to point something out?
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    I was trying to point out how the fact that complex roots come in conjugate pairs can be useful. We can also use it to prove that a real polynomial can be factorized into linear and quadratic real factors, although as you say it requires the fundamental theorem of algebra which is non-trivial.
 
 
 
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