Results 1 to 2 of 2

Math Help - Complex numbers

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    58

    Complex numbers

    Please help me to solve:

    Prove
    Rez>0 \; \: and \; \; z\neq 1 \Rightarrow  \left |\frac{z+1}{z-1}  \right |>1

    Thanks a lot
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by sinichko View Post
    Prove
    Rez>0 \; \: and \; \; z\neq 1 \Rightarrow  \left |\frac{z+1}{z-1}  \right |>1
    \left| z \right|^2  + 2\text{Re} (z) + 1 > \left| z \right|^2  - 2\text{Re} (z) + 1~~~[1]
    From the given it should be clear that [1] is true.

    Recall that  z + \overline z  = 2\text{Re} (z)

    You need to verify that \left| z \right|^2  + 2\text{Re} (z) + 1 = \left| z \right|^2  + z + \overline z  + 1 = \left( {z + 1} \right)\left( {\overline z  + 1} \right)
    and \left| z \right|^2  - 2\text{Re} (z) + 1 = \left| z \right|^2  - z - \overline z  - 1 = \left( {z - 1} \right)\left( {\overline z  - 1} \right)

    But \left( {z + 1} \right)\left( {\overline z  + 1} \right) = \left| {z + 1} \right|^2 \;\& \,\left( {z - 1} \right)\left( {\overline z  - 1} \right) = \left| {z - 1} \right|^2 which implies \left| {z + 1} \right|^2  > \left| {z - 1} \right|^2 .

    You can finish.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 03:14 PM
  3. Imaginary numbers/complex numbers
    Posted in the Algebra Forum
    Replies: 7
    Last Post: August 25th 2009, 11:22 AM
  4. Replies: 2
    Last Post: February 7th 2009, 06:12 PM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum