Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complexs equation

  1. Aug 27, 2011 #1
    I have a couple of problems with the conjugate. I have two equations to solve, (conjugate)z=2/z and (conjugate)z=-2/z
    How do I solve these= Are there some rules when you use the conjugate?
     
  2. jcsd
  3. Aug 27, 2011 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi matsorz! :smile:

    [itex]z + \bar{z} = 2Re(z)[/itex]

    [itex]z - \bar{z} = 2Im(z)[/itex]

    [itex]z\bar{z} = |z|^2[/itex] :wink:
     
  4. Aug 27, 2011 #3
    Ok, so I put |z^2|=2 in a) and |z^2|=-2 in b? Do I put in x and y, or do i just solve for z straight away?
     
  5. Aug 27, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    One problem you have is that the second equation is impossible.
    [tex]\overline{z}= -\frac{2}{z}[/tex]
    is the same as [itex]z\overline{z}= -2[/itex] but [itex]z\overline{z}= |z|^2[/itex] must be a positive real number.
     
  6. Aug 27, 2011 #5
    (a-bi)*(a+bi)=2
    a**2 + b**2 = 2

    center ( 0,0 ) radius =sqrt2

    (a-bi)*(a+bi)=-2

    center (a,b) radius = i sqrt2

    take abs of radii both

    one is sqrt2
    other abs ((sqrt 2) *i)=abs(i) * abs(sqrt2)

    abs(i)=1

    so they turn out to be equal.
     
  7. Aug 27, 2011 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    straight away …

    (it's always quickest to avoid coordinates as far as you can)

    |z| = √2 (the first case), so the general solution for z is … ? :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook