Results 1 to 7 of 7

Math Help - Complex Number roots Easy question!

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    28

    Complex Number roots Easy question!

    Hi I have this question that is very easy I am sure but I am having a hard time trying to figure out what the last part is asking! anyways here is the question:

    Find the four roots of z^4 + 4 = 0 and use them to factor z^4 + 4 into the product of two quadratic factors with real coefficients.

    Ok I was able to find the 4 roots with ease they are (1+i), (-1+i), (1-i), (-1-i)

    Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well

    My GUESS was (z^2-2z+2)(z^2+2z+2) but I didnt use the roots at all to come to that answer I used some formula I found on web.

    So please someone out there make me less dumb.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,570
    Thanks
    1428
    z^4 + 4 = 0

    z^4 = -4

    z^4 = 4\left(\cos{\pi} + i\sin{\pi}\right)

    z = 4^{\frac{1}{4}}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)

    z = \sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)


    Now it's important to note that all the roots are going to be evenly spaced around a circle, so they will all have a magnitude of \sqrt{2} and differ by angles of \frac{2\pi}{4} = \frac{\pi}{2}.

    Find the other roots.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    28
    did you read my entire post? I'm having trouble with the 2nd part of question. that has to do with product of two quadratic factors with real coefficients.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by fizzle45 View Post
    Hi I have this question that is very easy I am sure but I am having a hard time trying to figure out what the last part is asking! anyways here is the question:

    Find the four roots of z^4 + 4 = 0 and use them to factor z^4 + 4 into the product of two quadratic factors with real coefficients.

    Ok I was able to find the 4 roots with ease they are (1+i), (-1+i), (1-i), (-1-i)

    Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well

    My GUESS was (z^2-2z+2)(z^2+2z+2) but I didnt use the roots at all to come to that answer I used some formula I found on web.

    So please someone out there make me less dumb.
    Note that (z-(1+i))(z-(1-i)) = z^2-2z+2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by fizzle45 View Post
    Now for the part that asks product of two quadratic factors with real coefficients. this is where I Because I have no idea what they are even asking and my book/teacher expects us to already know this because its so basic... well
    If (x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0, the coefficients of x in f(x) are are a_{0}, a_{1}, a_{2}, ..., a_{n}, and could be real or complex numbers. For example, if we have p(x) = 3x^2+5x+7 and q(x) = 5ix^4+ix, then all the coefficients of x in p(x) are real (i.e., 3, 5, 7 are real) but the coefficients of q(x) are not (since 5i and i are complex). When they ask you to provide two quadratic factors with real coefficients, they want two quadratic functions whose coefficients are not complex numbers. If you have numbers n_{1},...., n_{k}, then the minimal polynomial for which all these numbers satisfy is (x-n_{1})\cdots(x-n_{k}). Take 1, 2, 3. To find the polynomial whose roots are these, we find the product (x-1)(x-2)(x-3) = (x^2-3x+2)(x-3) = x^3-6 x^2+11 x-6. Note that if we were asked to write it as a product of a linear and a quadratic, we would have just stopped at the second step. What you were asked is bit reversed but essentially the same: given the roots of f(x) = x^4+4 = 0 are a, b, c, and d (all of which you have found), write f(x) as a product of two quadratics with real coefficients.
    Last edited by TheCoffeeMachine; September 3rd 2010 at 09:35 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    The point of this exercise is to show you about conjugate pairs as roots of polynomials with real coefficients. If w is a complex number then \left( {z - w} \right)\left( {z - \overline w } \right) = z^2  - \left( {w + \overline w } \right)z + w\overline w  = z^2  - 2\text{Re} (w)z + \left| w \right|^2.
    That is a polynomial with real coefficients.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, fizzle45!

    Find the four roots of z^4 + 4 = 0

    and use them to factor z^4 + 4 into the product of two quadratic factors
    with real coefficients.

    Ok, I was able to find the 4 roots with ease.
    They are: (1+i), (1-i), (\text{-}1+i), (\text{-}1-i) .Good!

    Now for the part that asks product of two quadratic factors with real coefficients,
    I have no idea what they are even asking

    My GUESS was: (z^2-2z+2)(z^2+2z+2) . Right!
    but I didnt use the roots at all to come to that answer.
    I used some formula I found on web.

    If a polynomial equation,  f(z) = 0,has four roots: . a,b,c,d
    . . then the equation factors into: . (z-a)(z-b)(z-c)(z-d) \:=\:0


    We have the equaton: . z^4 + 4 \:=\:0
    . . with four roots: . (1+i),\:(1-i),\;(\text{-}1+i),\;(\text{-}1-i)


    So the equation factors:

    . . \bigg[z - (1+i)\bigg]\,\bigg[z - (1-i)\bigg]\,\bigg[z-(-1+i)\bigg]\bigg[z-(-1-i)\bigg] \;=\;0

    . . . . . \underbrace{\bigg[(z-1) - i\bigg]\,\bigg[(z-1) +i\bigg]}_{\text{multiply}}\,\underbrace{\bigg[(z+1) - i\bigg]\,\bigg[(z+1) + i\bigg]}_{\text{multiply}} \;=\;0


    The product of the first two factors is:
    . . (z-1)^2 - i^2 \:=\:z^2 - 2z + 1 + 1 \:=\:z^2-2z + 2

    The product of the last two factors is:
    . . (z+1)^2 - i^2 \:=\:z^2 + 2z + 1 + 1 \:=\:z^2 + 2z + 2


    And that is why: . z^4 + 4 \;=\;(z^2 - 2z + 2)(z^2 + 2z + 2)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex number easy inequality
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: February 3rd 2010, 05:14 PM
  2. Complex Number Division. Easy
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 16th 2009, 03:47 PM
  3. [SOLVED] easy complex number questions
    Posted in the Algebra Forum
    Replies: 8
    Last Post: October 23rd 2009, 11:20 PM
  4. Complex number roots
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 5th 2009, 07:05 AM
  5. complex number roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 25th 2008, 07:28 PM

Search Tags


/mathhelpforum @mathhelpforum