1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Whats wrong here?

  1. May 1, 2006 #1
    [tex]sin(z)=2[/tex]
    [tex]e^{iz}-e^{-iz} = 4i[/tex]
    [tex]e^{2iz}-4ie^{iz} = 1[/tex]
    [tex]iz \ln (e^{iz}-4i) = 0[/tex]
    [tex] z=0[/tex]

    when solving it by
    [tex]w = e^{iz}[/tex]
    [tex]w^2-4wi-1 = 0[/tex]
    i get one more solution, why is the first way not as good as the second way?
     
  2. jcsd
  3. May 1, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ln(ab)= ln(a)+ ln(b), not ln(a)*ln(b).

    If [tex]e^{2iz}-4ie^{iz} = 1[/tex]
    then [tex]e^{iz}(e^{iz}- 4i)= 1[/tex]
    so [tex]iz+ ln(e^{iz}- 4i)= 0[/tex]
    NOT [tex]iz \ln (e^{iz}-4i) = 0[/tex]
     
  4. May 2, 2006 #3
    hehe, right, that was a dumb mistake :biggrin:
    thank you for pointing it out.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?