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Simple questions about complex numbers

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data

    1) If I know that z_1 = (2+i) and z_2 = (2-i) are solutions to a polynomial, how do I find it? (I have six to chose between, it's a multiple choice). Do I just insert and see of it equals zero?

    2) When I know that z^2 has modulus 4 and argument pi/2, how do I find modulus and argument for z?

    3) If I know a polynomial (real numbers, not complex!) f(x) = 2x^2 and g(x) = cx, how do I find the number c, so g(x) is a tangent to the polynomial?

    3. The attempt at a solution

    1) I would just insert solutions.

    2) I do not know how to approach this problem.

    3) First, g(x) = f(x), and the gradient of g(x) must equal f'(x) in the point x_0?
  2. jcsd
  3. Sep 1, 2007 #2


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    Dearly Missed

    1) Quite correct, although there might be additional clues in the actual polynomials given that makes the determination process easier.

    2. You have:
    [tex]z^{2}=4e^{i\pi(\frac{1}{2}+2n)}, n\in{Z}[/tex]
    Take the square root of this and see how many z's fits the expression.

    3) I don't get your argument f(x)=g(x)?
  4. Sep 1, 2007 #3
    1) Cool, thanks.

    2) So modulus and argument for z is: mod = 2 and argument = pi/4?:

    w_0 = 4^(1/2)*e^i*(pi/2)/2)

    I don't have to find w_1, eh?

    3) Nevermind, a silly question. I just had to read a section in my book.
  5. Sep 1, 2007 #4


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    There is another argument to consider for 2)...
  6. Sep 1, 2007 #5
    pi/4 + 2kpi?
  7. Sep 1, 2007 #6


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    Nope... that's the same angle. :wink: But you're close.
  8. Sep 1, 2007 #7
    D'oh :-)

    I hate to guess, but is it something like pi/4 +(2pi/n)*k = pi/4 + pi?
  9. Sep 1, 2007 #8


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    You got the right answer... but I'm kind of unsure about your left-handside formula... I'd do it like this using arildno's idea... you know that [tex]2\theta = \pi/2 + 2n\pi[/tex] then dividing by 2, [tex]\theta = \pi/4 + n\pi[/tex]... you only need to consider n = 0 and n = 1... which give [tex]\pi/4[/tex] and [tex]5\pi/4[/tex]... all the other n's repeat the same angles.
  10. Sep 1, 2007 #9
    Great, thanks!
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