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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


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    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


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    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]
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