Proving E = 1: Exploring Complex Numbers

[nameonascreen]

I saw this the other day and I don't understand how complex numbers work well enough to disprove it:

e^(2*pi*i) = 1

e^(2*pi*i)^(1/(2*pi*i)) = 1^(1/(2*pi*i))

Then the left side equals e and the right side equals 1.

 

[arildno]

Perhaps you ought to learn enough about complex numbers so that you don't make a wild goose chase trying to disprove a provable true statement.

Your error lies in assuming that the rules for exponentiation that are valid for real numbers are valid for complex number.

They are not; rather, real number exponentiation is a special case of how exponentiation works in the complex number plane.

 

[nameonascreen]

so e = 1 in a complex number plane?

 

[Mentallic]

No. What he is trying to say is that the rule for exponentiation $$\left(a^b\right)^c=a^{bc}$$ is only valid when dealing with real numbers, $$a\geq 0$$. When you extend into the complex plane, this rule breaks down.

 

[nameonascreen]

ok cool. thanks! so what really happens when you raise e^(2*pi*i) = 1 to the power of 1/(2*pi*i)?

 

[HallsofIvy]

ok cool. thanks! so what really happens when you raise e^(2*pi*i) = 1 to the power of 1/(2*pi*i)?

Since $2\pi$ is irrational, there will be an infinite number of values. One of them, of course, will be 1.

In general, to find $a^r$ for general complex numbers you use $a^r= e^{ln(a^r)}= e^{r ln(a)}$.

Here, because $1/(2\pi i)= -i/(2\pi)$ that would be
$$e^{-\frac{i}{2\pi}ln(1)}$$

Now, restricted to the real numbers, ln(1)= 0 so that would give $e^0= 1$. But in the complex numbers, $ln(1)= 2k\pi i$ for any integer k.

That is,
$$1^{\frac{1}{2\pi i}}= e^{-\frac{i}{2\pi}2k\pi i}= e^k$$
for any integer k. Of course, k=0 gives 1, and k= 1 gives e, your two values.