This shows how (1+i * Pi/N)^k, k=1..N traces out a semi-circle for large values of N.
Geometrically, all it says is:
* Draw a right triangle ABC with AB=1, BC=Pi/N, and ABC the right angle
* Make a copy of ABC, call it A'B'C', and scale it so that A'B'(the long leg) = AC (the hypotenuse)
* Put A'B'C' over ABC so that A'B' and AC coincide
* Let ABC=A'B'C'
* Repeat the process N times
* Look where you end up when N is large enough
The answer is: when N is large, Pi/N is small, and the right triangle ABC is almost isosceles, AB ~= AC. So you end up with N slices of a pie that make up a fraction of a circle.
Which fraction? Well, the perimeter is N/Pi * N = Pi - so half a circle. So if A=(0,0) and B=(1,0), you end up at (-1,0).
Now (1+x/n)^n approaches e^x, so it makes sense to define e^(i * Pi) to be the same limit - which we found out to be -1 + i * 0.
My favorite angle on this is the following graphic/animation, also present in the thread:
https://upload.wikimedia.org/wikipedia/commons/0/0e/ExpIPi.g...
This shows how (1+i * Pi/N)^k, k=1..N traces out a semi-circle for large values of N.
Geometrically, all it says is:
* Draw a right triangle ABC with AB=1, BC=Pi/N, and ABC the right angle
* Make a copy of ABC, call it A'B'C', and scale it so that A'B'(the long leg) = AC (the hypotenuse)
* Put A'B'C' over ABC so that A'B' and AC coincide
* Let ABC=A'B'C'
* Repeat the process N times
* Look where you end up when N is large enough
The answer is: when N is large, Pi/N is small, and the right triangle ABC is almost isosceles, AB ~= AC. So you end up with N slices of a pie that make up a fraction of a circle.
Which fraction? Well, the perimeter is N/Pi * N = Pi - so half a circle. So if A=(0,0) and B=(1,0), you end up at (-1,0).
Now (1+x/n)^n approaches e^x, so it makes sense to define e^(i * Pi) to be the same limit - which we found out to be -1 + i * 0.