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 [tex]e^\frac{i\pi}{2} = 1 [/tex], it seems as though [tex]\frac{i\pi}{2} = 0[/tex]. 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 [tex]i\pi = 0[/tex] which would in turn imply that [tex]e^{i\pi} = 1[/tex] 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 [tex]\ln(1)[/tex] 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%27s_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 [tex]e^{2i\pi} = 1[/tex] 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 [tex]e^{2i\pi} = 1[/tex] which leads to essentially the same question I had originally.

acabus

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

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

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

D H

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 [itex]\exp(v) = z[/itex] as the "principal value" [itex]v=\textrm{Log}\,z[/itex]. The typical choice is that [itex]-\pi < \arg v \le \pi[/itex]. This choice makes [itex]\textrm{Log}\,1 = 0[/itex], which is the obvious choice. But what about [itex]\textrm{Log}\,(-1)[/itex]? Given Euler's identity [itex]\exp \pi i + 1 = 0[/itex], one "obvious" choice is [itex]\textrm{Log}\,(-1) = \pi i[/itex].

There's a minor problem here. For small real ε > 0, this choice makes [itex]\textrm{Log}(-1+\epsilon i) \approx \pi i[/itex] but it makes [itex]\textrm{Log}(-1-\epsilon i) \approx -\pi i[/itex]. 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

Euler's formula is

[itex]e^{ix} = cos(x) + i\ sin (x)[/itex]

With that formulation, it's easy to see that [itex]e^{ix}[/itex] 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:
[tex]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[/tex]
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:
[tex]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[/tex]
Now remember that i² is -1, so we substitute accordingly:
[tex]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[/tex]
Hmm, what happens if we split this into two components, so that e^bi=f(b)+i*g(b). We get:
[tex]f(b)=1-\frac{b^{2}}{2!}+\frac{b^{4}}{4!}-\frac{b^{6}}{6!}+\ldots[/tex]
[tex]g(b)=b-\frac{b^{3}}{3!}+\frac{b^{5}}{5!}-\frac{b^{7}}{7!}+\ldots[/tex]
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=[itex]\pi[/itex] and we get e^{[itex]\pi[/itex]i}=cos([itex]\pi[/itex])+i*sin([itex]\pi[/itex])=-1.

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