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

Complex Numbers & Identities

  1. May 3, 2007 #1
    I was never good at trigonometric identities.


    Let z= cos x + i sin x

    Express 2/(1 + z) in the form 1 - i tan kx



    I need help. A pointer to where to start would be great.
     
  2. jcsd
  3. May 3, 2007 #2
    [tex]z=e^{ix}[/tex]

    and
    [tex]\tan(kx)=\frac{1}{2i}\cdot\frac{e^{ikx}-e^{-ikx}}{e^{ikx}+e^{ikx}}[/tex]
     
  4. May 3, 2007 #3
    I'm sorry, but I have never seen that identity before. It's not in my math tables?????
     
  5. May 3, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you know that [itex]tan z= \frac{sin z}{cos z}[/itex]? Replace sin z with [itex](e^{ix}-e^{-ix})/2i[/itex] and cos z with [itex](e^{ix}+ e^{-ix})/2[/itex]

    I would have been inclined to do this problem more directly. Write 2/(1+ z) as
    [tex]\frac{2}{1+ cos x+ i sin x}= \frac{2}{(1+cos x)+ i sin x}[/tex]
    and multiply both numerator and denominator by the "complex conjugate", (1+ cos x)- i sin x.
     
  6. May 4, 2007 #5

    Gib Z

    User Avatar
    Homework Helper

    I'm sue Joza hasnt seen polar form yet...HallsofIvys method is the best way for you.
     
  7. May 4, 2007 #6
    No I understand polar form. And yes I understand that tan is sin/cos. I haven't done 3rd level mathematics yet though.
     
  8. May 4, 2007 #7

    Gib Z

    User Avatar
    Homework Helper

    O well thats good then, Halls's method is still the way to go though.
     
  9. May 4, 2007 #8

    uart

    User Avatar
    Science Advisor

    That seemed the most obvious way to proceed to me as well. I was going to post exactly that but Tim beat me to it, then I decided I liked his way better.

    Here's the reason. If you follow through with the "conjugate" method and simplify as much as possible you end up with,

    [tex] \frac{2}{1+z} = 1 - i \, \frac{\sin(x)}{1+\cos(x)} [/tex].

    While this is tantalizingly close to the form that OP requires it's still not quite there. To finish off you need to be able the see that you can use the following two trig identities (on the numerator and denominator respectively),

    [tex] \sin(x) = 2 \sin(x/2) \cos(x/2) [/tex]

    and

    [tex] 1 + \cos(x) = 2 \cos^2(x/2) [/tex].

    This last step wasn't too obvious to me, actually I didn't even see it at first. It was only after working through Tim’s method that I realized that "k" had to be equal to 1/2 and then the half angle substitutions idea came to me.
     
    Last edited: May 4, 2007
  10. May 4, 2007 #9
    actually... just look at
    [tex]1-\frac{2}{z+1}=\frac{z-1}{z+1}[/tex] alone with the identity will give you everything.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Complex Numbers & Identities
  1. Complex number (Replies: 6)

  2. Complex numbers? (Replies: 6)

  3. Complex numbers (Replies: 2)

  4. Complex numbers (Replies: 2)

Loading...