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

[Hacky]

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

 

[d_leet]

The easiest way to at least see this would probably be to factor i out of the numerator.

 

[robphy]

Note that $$z=a+bi = re^{i\theta}$$ and $$\bar z=a-bi = re^{-i\theta}$$.

So, $$\frac{z}{\bar z}=e^{i(2\theta)}$$.

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

 

[HallsofIvy]

Or, the standard way to represent a fraction as a complex number: multiply both numerator and denominator by the complex conjugate of the denominator:
$$\frac{i- 1}{i+ 1}= \frac{i-1i}{i+ 1}\frac{-i+1}{-i+1}$$
$$= \frac{(i-1)(-i+1)}{1- i^2}= \frac{-i^2+ 2i+ 1}{1+1}= \frac{2i}{2}= i$$