# Sinx and i

1. Feb 12, 2004

### Gunni

Quick question.

$$sin(-x) = -sin(x)$$

this can be seen as this example:

$$sin(i^2x) = i^2sin(x)$$

Does this then apply?

$$sin(ix) = isin(x)$$ ?

I'm trying to derive a formula for the length of a simple parabola. Unfortunetly in my calculations I end up with Arcsin(i2x) along the way and it would make my life alot simpler if that could be treated as iArcsin(2x). Then the i would be divided out and then I'd get a real number instead of a complex number.

It's sort of a dare, I'm taking the calculus course where we learn about complex numbers but I read the book during the christmas vacation so I've got most of the basics down. Right now we're learning about the length of functions and we asked if our teacher would give us an example using a parabola of the form Ax^2 + C, but she said it was too complex. Of course, once she says that we spend the rest of the week trying to do it by ourselfes.

Thanks a lot.

Last edited: Feb 12, 2004
2. Feb 12, 2004

### Dr Transport

Try plugging ix into the definition of the sin function (e^ix-e^(-ix))/2i. You'll find that you get sin(ix) = i*sinh(x) not i*sin(x).

3. Feb 12, 2004

### Gunni

Isn't that the definition of the sinh function? I always thought sin was defined as a/c in a triangle with one 90° corner.

4. Feb 12, 2004

### chroot

Staff Emeritus
5. Feb 12, 2004

### matt grime

absolutely not

What;s the length of a function?

6. Feb 12, 2004

### Gunni

$$L = \int_{a}^{b}\sqrt(1 + (f'(x))^2)dx$$

Anyway, thanks guys. Too bad that doesn't apply, seems the damn teacher was right.

7. Feb 12, 2004

### NateTG

Re: Re: Sinx and i

Since this is calculus I would guess that he's talking about the arclength. In complex analysis it's more likely to be a line integral.

Of course, the arc length of a parabola is IIRC not so bad except for the tricky integral:

$$\int_a^b \sqrt{\frac{dy}{dx}^2+1}dx$$
so
$$\int_a^b \sqrt{4x^2+1} dx$$
which leads to the naive substitution
$$x(u)=\frac{i\sin(u)}{2}$$
so
$$\int_a^b cos(u) dx$$
but it needs to be inverted to get the correct limits of integration:
$$\int_{-\sin^{-1}(2ia)}^{-\sin^{-1}{2ib}} (1+\frac{i}{2}) \cos(u) du$$
$$(1+\frac{i}{2}) |_{-\sin^{-1}(2ia)}^{-\sin^{-1}{2ib}} -sin u$$
so
$$(1+\frac{i}{2})(2b-2a)i$$
resulting in the arc length:
$$(a-b)+i2(b-a)$$

Unfortunately this is a less than ideal result. Perhaps he should try the substitution:
$$x=\frac{1}{2}\tan(u)$$

Last edited: Feb 12, 2004
8. Feb 12, 2004

### matt grime

that isn't the 'length of a function'. it is the arc length. is your teacher really talking about lengths of functions?

as nate points out it is realtively easily to find in the case of a parabola.

9. Feb 12, 2004

### Gunni

I'm sorry, I'm from Iceland and math is tought in icelandic, so I don't know all the english names for functions.

I'll look over what NateG posted tomorrow (kind of late here at GMT), since I've run into trouble deriving it myself the least I can do is learn how to do it properly. Thanks guys.