Question regarding imaginary numbers and euler's formula

[elvishatcher]

So, I was thinking about Euler's formula, and I noticed something interesting. Based on the fact that e^\frac{i\pi}{2} = 1, it seems as though \frac{i\pi}{2} = 0. However, this doesn't make any sense. Not only can I not see how this expression could possibly equal 0, but that would imply that i\pi = 0 which would in turn imply that e^{i\pi} = 1 when it, of course, is equal to -1. At the moment, I have only a very basic understanding of complex/imaginary numbers and their properties, but it seems to me that the implication here is that \ln(1) is not uniquely equal to zero. Is there something I'm missing that shows that this is not the case? If I am correct in this conclusion, is this because of some property of imaginary numbers that I don't know about yet?

 

[jedishrfu]

e ^ (i * pi / 2) = i

not 1

based on eulers formula cos x + i sin x = e ^ i x so with pi /2 we get cos pi /2 = 0 and sin pi/2 = 1

hence the answer i

http://en.wikipedia.org/wiki/Euler's_formula

 

[elvishatcher]

Oh dear, how embarassing. What a silly mistake.

However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that e^{2i\pi} = 1 which leads to essentially the same question I had originally.

 

[elvishatcher]

Oh dear, how embarassing. What a silly mistake.

However, now that I think about it, while that may have been a silly arithmetic mistake, it is true that e^{2i\pi} = 1 which leads to essentially the same question I had originally.

 

[acabus]

Euler's formula says that e^{ix} = \cos x + i \sin x.

Let x = 2 \pi n, where n is any integer. Then e^{ix} = \cos x + i \sin x = 1 + 0i = 1.

Therefore, e^{ix} = 1 has an infinite number of solutions, all of the form x = 2 \pi n. The two you brought up, x=0 and x = 2 \pi are just the ones where n = 0 and 1 respectively.

 

[D H]

I have only a very basic understanding of complex/imaginary numbers and their properties, but it seems to me that the implication here is that \ln(1) is not uniquely equal to zero.

Correct. The complex logarithm is a multi-valued function. (Aside: "multi-valued function" is a bit of a misnomer, as is "red herring". Red herrings typically are neither red nor are they fishes. Multi-valued functions such as the inverse sine and the complex logarithm are not "functions".)

When you learn a bit more about complex analysis you'll run into the concept of branch points, branch cuts, and principal values. To make the complex logarithm a true function, one has to (somewhat arbitrarily) choose exactly one of the infinite number of values v that satisfy \exp(v) = z as the "principal value" v=\textrm{Log}\,z. The typical choice is that -\pi < \arg v \le \pi. This choice makes \textrm{Log}\,1 = 0, which is the obvious choice. But what about \textrm{Log}\,(-1)? Given Euler's identity \exp \pi i + 1 = 0, one "obvious" choice is \textrm{Log}\,(-1) = \pi i.

There's a minor problem here. For small real ε > 0, this choice makes \textrm{Log}(-1+\epsilon i) \approx \pi i but it makes \textrm{Log}(-1-\epsilon i) \approx -\pi i. There's a discontinuity in the neighborhood of z=-1 (or in the neighborhood any other negative real number). That's what you get with branch cuts.

 

[elvishatcher]

Very interesting - thanks for the info!

 

[SteveL27]

Very interesting - thanks for the info!

Euler's formula is

e^{ix} = cos(x) + i\ sin (x)

With that formulation, it's easy to see that e^{ix} must be periodic with period 2pi. Once you get used to thinking about the complex exponential that way, it gets much easier to visualize.

 

[elvishatcher]

Ah, I hadn't thought about the periodicity before. Thanks

 

[Eval]

The cool proof that I saw goes like this:
Take the Maclaurin series for e^u. That is:
e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\frac{u^{4}}{4!}+\frac{u^{5}}{5!}+\frac{u^{6}}{6!}+\frac{u^{7}}{7!}+\ldots
Now, say u=a+bi. Then e^u=e^a+bi=e^ae^bi. We are interested in the e^bi part, so let's see what that comes out to:
e^{bi}=1+(bi)+\frac{(bi)^{2}}{2!}+\frac{(bi)^{3}}{3!}+\frac{(bi)^{4}}{4!}+\frac{(bi)^{5}}{5!}+\frac{(bi)^{6}}{6!}+\frac{(bi)^{7}}{7!}+\ldots
Now remember that i² is -1, so we substitute accordingly:
e^{bi}=1+bi-\frac{b^{2}}{2!}-\frac{ib^{3}}{3!}+\frac{b^{4}}{4!}+\frac{ib^{5}}{5!}-\frac{b^{6}}{6!}-\frac{ib^{7}}{7!}+\ldots
Hmm, what happens if we split this into two components, so that e^bi=f(b)+i*g(b). We get:
f(b)=1-\frac{b^{2}}{2!}+\frac{b^{4}}{4!}-\frac{b^{6}}{6!}+\ldots
g(b)=b-\frac{b^{3}}{3!}+\frac{b^{5}}{5!}-\frac{b^{7}}{7!}+\ldots
Wait! Now, f(b) is the Mclaurin series for cos(b) and g(b) is the Mclaurin series for sin(b). We have, then, that e^bi=f(b)+i*g(b)=cos(b)+i*sin(b). Now, plug in b=\pi and we get e^\pii=cos(\pi)+i*sin(\pi)=-1.

I find the computation pretty cool. It is a testament to the power of infinite series :)