^{iπ}= -1. Here

*i*is the complex number, the square root of -1.

This is a special case of the more general

e

^{iθ}= cos θ +*i*sin θ
Thus, not only is e

So far, so good.

^{iπ}= -1, but e^{i2π}= +1.So far, so good.

Let's evaluate the following: (e

On the one hand, since e

^{i2π})^{π}. That is, the quantity e^{i2π}raised to the π power.On the one hand, since e

^{i2π}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 (e

^{i2π})^{π}= e^{i2π2}. Now using the formula given above, this quantity is cos(2π^{2}) +*i*sin(2π^{2}), 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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