Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Identity****Member**- Registered: 2007-04-18
- Posts: 934

I'm trying to get the integral of

by factoring, partial fractions, and finally integration.Here's where I'm up to so far

Now I have a feeling that this approach *should* work. I've been using complex numbers a lot recently and they've never failed me in all kinds of ridiculous tasks. So where do I go from here, if it is possible to go from here?

Thanks

*Last edited by Identity (2010-05-16 01:49:51)*

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

If u can express your already complex expression in the Polar Form i.e. R.e[sup]iθ[/sup], then you can probably take the Complex Natural Log and change it to form logR+iθ.

I'm not sure if it will help you Simplifying your problem any further...

If two or more thoughts intersect, there has to be a point!

Offline

**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Thanks ZHero, I'm not sure what to do with the absolute value signs though. It acts as the modulus doesn't it? And everytime I evaluate the modulus of the stuff it comes out to be 1

Offline

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Yes! It does! Coz log of -ves are "Not Defined" (you can't blame Napier for that, except for, in case, he was Napping )!

Try entering "alternate <space> form <space> **your log expression**" into WolframAlpha and look at various Alternative Representations!

If two or more thoughts intersect, there has to be a point!

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi identity;

You have taken a wrong turn somewhere that answer is not simplifed at all . Please rework the problem. No help from wolfram alpha. I do not think that is the right answer.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**Identity****Member**- Registered: 2007-04-18
- Posts: 934

lol bobbym this is just some fun. I know the answer is

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Identity wrote:

Thanks ZHero, I'm not sure what to do with the absolute value signs though. It acts as the modulus doesn't it? And everytime I evaluate the modulus of the stuff it comes out to be 1

There are no "absolute values" in complex analysis, only modulus.

The modulus of z is 1, as you said. This has to happen, otherwise you'll get a nonreal result. Now just calculate the arg(z) by finding the real and imaginary parts, z = a + bi, and then arg(z) = arctan(b/a).

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi identity;

Well you got me! I was wondering why you expanded the denominator when it already was in perfect form for you. I know you know better than that.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Pages: **1**