Results 1 to 4 of 4

Math Help - Finding roots - Rouche's - calculating absolute value

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    95

    Finding roots - Rouche's - calculating absolute value

    I'm trying to find roots to a polynom using Rouche's theorem.

    Rouche's
    \displaystyle \left| f(z) - g(z) \right| < \left| f(z)  \right|-\left|  g(z) \right|

    I'm feeling unsure of how to calculate the absolute values, if anyone could help...

    I'll try an example, taken from wikipedia.

    z^5+3z+7 let f(z) = z^5 and g(z) = 3z + 7

    Examine |z| = 1
    |f(z)| = |z^5| = |z|^5 = |1|^5 = 1

    |g(z)| = |3z+7| = |7 - (- 3z)| = |7|-|-3z| = 7 - 3z = 4

    |f(z)| - |g(z)| = 1 - 4 = -3

    |f(z) - g(z)| = |z^5 -3z -7| = |1 - 3 - 7| = |-9| = 9

    If I've done this correctly, can I deduce that there's no roots in |z| = 1? If not, how would I show that there's no roots in |z| = 1?

    Comments
    Please point out errors in my calculation. Furthermore, if I examine |z| = 2 I get |f(z)| - |g(z)| = 31 and |f(z) - g(z)| = 19 which indicates that there are 5 roots inside |z| = 2. Or..?
    Last edited by liquidFuzz; December 3rd 2010 at 08:17 AM. Reason: bad spelling
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Let's see, I don't understand why do you use the difference, we have:

    F(z)=7+(z^5+3z)=f(z)+g(z) ( f,g analytical functions on |z|\leq 1 )

    On |z|=1 :

    |f(z)|=7, |g(z)|=|z^5+3z|\leq |z|^5+|3z|=4

    That is:

    |f(z)|>|g(z)| on |z|=1

    According to Rouche's theorem, the number of roots of F(z) on |z|<1 is equal to the number of roots of f(z) on |z|<1 (including multiplicities).

    As a consequence, F(z) has no roots on |z|<1 .

    Regards.

    Fernando Revilla
    Last edited by FernandoRevilla; December 3rd 2010 at 12:17 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    95
    Ok, thanks!

    Did you assume no roots while setting f(z) g(z)?

    I like to try inside |z| = 2 please help if needed.

    f(z) = z^5 - has got 5 roots
    g(z) = 3z + 7

    \left|f(z)\right| = \left|z^5\right| = \left|z\right|^5 = 2^5
    \left|g(z)\right| = \left|3z+7\right| = \left|7-(-3z)\right| = 7-6 = 1

    So, |f(z)| > |g(z)| thus z⁵ + 3z + 7 has got 5 roots in |z| = 2. Or..?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by liquidFuzz View Post
    So, |f(z)| > |g(z)| thus z⁵ + 3z + 7 has got 5 roots in |z| = 2. Or..?
    Right. On |z|=2 :

    |z^5|>|3z+7|

    So,

    F(z)=z^5+3z+7 has 5 roots in |z|<2

    Regards.

    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Properties of Roots and Absolute Values
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 6th 2011, 12:44 PM
  2. Square Roots turning into Absolute Value
    Posted in the Algebra Forum
    Replies: 5
    Last Post: June 23rd 2011, 04:56 PM
  3. Rouche's Theorem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 14th 2011, 12:39 PM
  4. Rouché's theorem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 12th 2011, 01:31 PM
  5. finding absolute maximum and absolute minimum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2009, 05:22 PM

Search Tags


/mathhelpforum @mathhelpforum