Results 1 to 6 of 6

Math Help - complex no

  1. #1
    Senior Member
    Joined
    Feb 2010
    Posts
    456
    Thanks
    34

    complex no

    if z1,z2,z3...zn lie on circle mod(z)=2 then the value of mod(z1+z2+...+zn)-4*mod(1/z1+1/z2+1/z3...+1/zn) is
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093

    Re: complex no

    is the question

    \mid z_1+z_2+...+z_n -4 \mid \cdot \mid \frac{1}{z_1} + \frac{1}{z_2} + ...+\frac{1}{z_n}\mid

    or


    \left(\mid z_1+z_2+...+z_n  \mid -4 \right) \cdot \mid \frac{1}{z_1} + \frac{1}{z_2} + ...+\frac{1}{z_n}\mid
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2010
    Posts
    456
    Thanks
    34

    Re: complex no

    complex no-codecogseqn-4-.gif
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: complex no

    Quote Originally Posted by prasum View Post
    if z1,z2,z3...zn lie on circle mod(z)=2 then the value of
    |(z1+z2+...+zn)|-4|(1/z1+1/z2+1/z3...+1/zn)| is
    \left| {\sum\limits_{k = 1}^n {z_k } } \right| - 4\left| {\sum\limits_{k = 1}^n {\frac{1}{{z_k }}} } \right| = 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093

    Re: complex no

    Quote Originally Posted by Plato View Post
    \left| {\sum\limits_{k = 1}^n {z_k } } \right| - 4\left| {\sum\limits_{k = 1}^n {\frac{1}{{z_k }}} } \right| = 0

    I know that

     \mid z_1 + z_2 + ... +z_n \mid\;\; \leq\; \mid z_1 \mid\; + \mid z_2 \mid\; + ...+\mid z_n\mid = 2n

    \left| \frac{1}{z_1} +\frac{1}{z_2} + ...+ \frac{1}{z_n} \right| \;\; \leq\; \left| \frac{1}{z_1}\right|\; + \;\left| \frac{1}{z_2} \right| \; + ...+\;\left| \frac{1}{z_n}\right| = \frac{n}{2}

    you consider the equally without the bigger why ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2011
    From
    Awetuouncsygg
    Posts
    182
    Thanks
    12

    Re: complex no

    Let S=\sum_{k=1}^{n} z_k.

    |z_k|=2\Rightarrow z_k\cdot \overline{z_k}=4\Leftrightarrow \frac{1}{z_k}=\frac{\overline{z_k}}{4}

    4\left | \sum_{k=1}^{n} \frac{1}{z_k} \right |=\left | \sum_{k=1}^{n} \frac{4}{z_k} \right | =\left | \sum_{k=1}^{n} \overline{z_k} \right |=\left | \overline{S}\right |=\left | S\right |

    * That inequalities aren't really useful, if you multiply by -4 the second one it becomes -4\left |\sum_{k=1}^{n} \frac{1}{z_k} \right| \geq -2n (so you can't add them).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum