- #1

- 23

- 0

- Thread starter elvishatcher
- Start date

- #1

- 23

- 0

- #2

jedishrfu

Mentor

- 12,392

- 6,152

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

- #3

- 23

- 0

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.

- #4

- 23

- 0

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.

- #5

- 45

- 0

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.

- #6

- 15,393

- 685

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".)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.

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

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.

Last edited:

- #7

- 23

- 0

Very interesting - thanks for the info!

- #8

- 795

- 7

Euler's formula isVery interesting - thanks for the info!

[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.

- #9

- 23

- 0

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

- #10

- 33

- 0

Take the Maclaurin series for e

[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

[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 i

[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

[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

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

- Last Post

- Replies
- 14

- Views
- 7K

- Last Post

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 11

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 4K

- Replies
- 1

- Views
- 2K

- Replies
- 5

- Views
- 4K

- Last Post

- Replies
- 6

- Views
- 958

- Replies
- 2

- Views
- 4K

- Last Post

- Replies
- 13

- Views
- 4K