This is a special case of the more general
eiθ = cos θ + i sin θ
Thus, not only is eiπ = -1, but ei2π = +1.
So far, so good.
So far, so good.
Let's evaluate the following: (ei2π)π. That is, the quantity ei2π raised to the π power.
On the one hand, since ei2π is equal to 1, this is just 1 raised to the π power, which is still just one.
On the one hand, since ei2π is equal to 1, this is just 1 raised to the π power, which is still just one.
On the other hand, using simple algebra (power times exponent) we can write (ei2π)π= ei2π2. Now using the formula given above, this quantity is cos(2π2) + i sin(2π2), which is most assuredly not equal to one.
What gives? Where did error creep in?
One raised to any power is not "still just one".
ReplyDeleteFor fractional powers (p), things get tricky. For example, 1^(1/2) (i.e., the square root of 1), could be 1, or it could be -1. Similarly, 1^(1/4) could be i, or -i, or 1 or -1.
The formula cos(2\pip) + i sin(2\pip) gives one of these solutions in both of these cases [i.e., cos(\pi) + i sin(\pi) = -1 and cos(\pi/2) + i sin(\pi/2) = i] and provides a solution in the case of p=\pi
Ahh... I've been found out!
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