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

Why is thes true: i=(i-1)/(i+1)

  1. Oct 11, 2006 #1
    Of course I can see that i*(i+1)=(i-1) but is there some way (long division?) to show this in general? To show for example that i = {(2+i)(3+i)/(2-i)(3-i)}. Or to come up with these equivalencies, does one just multiply i by whatever you desire in the later expansion. I am reading about Schellbach's formulae to calculate Pi from i.

    Thanks, Howard
     
  2. jcsd
  3. Oct 11, 2006 #2
    The easiest way to at least see this would probably be to factor i out of the numerator.
     
  4. Oct 11, 2006 #3

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Note that [tex]z=a+bi = re^{i\theta}[/tex] and [tex]\bar z=a-bi = re^{-i\theta}[/tex].

    So, [tex]\frac{z}{\bar z}=e^{i(2\theta)}[/tex].

    In your two examples, (i-1) and (2+i)(3+i) make an angle of pi/2 with their complex conjugates.
     
  5. Oct 15, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Or, the standard way to represent a fraction as a complex number: multiply both numerator and denominator by the complex conjugate of the denominator:
    [tex]\frac{i- 1}{i+ 1}= \frac{i-1i}{i+ 1}\frac{-i+1}{-i+1}[/tex]
    [tex]= \frac{(i-1)(-i+1)}{1- i^2}= \frac{-i^2+ 2i+ 1}{1+1}= \frac{2i}{2}= i[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?