## Question regarding imaginary numbers and euler's formula

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?
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 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
 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.

## Question regarding imaginary numbers and euler's formula

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.
 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.

Mentor
 Quote by elvishatcher 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.
 Very interesting - thanks for the info!

 Quote by elvishatcher 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.
 The cool proof that I saw goes like this: Take the Maclaurin series for eu. 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 eu=ea+bi=eaebi. We are interested in the ebi 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 ebi=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 ebi=f(b)+i*g(b)=cos(b)+i*sin(b). Now, plug in b=$\pi$ and we get e$\pi$i=cos($\pi$)+i*sin($\pi$)=-1. I find the computation pretty cool. It is a testament to the power of infinite series :)