# Homework Help: Complex variables: Logarithm function.

1. Oct 2, 2008

### student85

I'm taking a complex variables course, and I'm really stuck at it, I've never felt this way in any math course before :S, I'm starting to get angry. Anyway here is the problem, I hope someone can give me a hand. I believe this is a basic and simple problem in the subject...

1. The problem statement, all variables and given/known data
Let D be the domain obtained by deleting the ray {x:x$$\leq$$0} from the plane, and let G(z) be a branch of log z on D. Show that G maps D onto a horizontal strip of width of 2$$pi$$
{x+iy: -$$\infty$$<x<$$\infty$$, co<y<co+2$$pi$$},
and that the mapping is one-to-one on D.

2. Relevant equations

3. The attempt at a solution
Ok so first off I'm trying to actually understand the problem. Where it says deleting the ray {x:x$$\leq$$0} I don't know exactly what it means. I mean, in my mind there is an infinite number of rays that satisfy those conditions (all the rays going from the imaginary axis and going left all the way to infinity, parallel to the x axis). I think I'm way off here, but believe me, I've read the textbook and it just isn't clear to me this all thing about rays and branches. Plus, I can't picture the branch of log z if I can't picture D in the first place! Could somebody help me with understanding this please?

Thanks a lot.

NOTE: The infinities are not supposed to be exponentials, I don't know why they appeared that way.

Last edited: Oct 2, 2008
2. Oct 2, 2008

### Dick

"Deleting the ray" just means don't define the log on it and don't expect the definition to be continuous when you cross that ray. The point to 'defining a branch' is to make the definition of the function continuous in the part of the domain that wasn't removed. Do you see why there would be a problem defining a single valued log if the domain contained a circle around zero?

3. Oct 2, 2008

### student85

Oooh, I think I know where you're getting. So if we have a complete circle around zero then the domain would be continuous and would go round and round... so, you'd get different values for G(z) as we added 2pi to z. Great great, I think I'm getting somewhere. I still have this question though: Particularly speaking about this problem now, what ray exactly is deleted with the ray {x:x$$\leq$$0} ? Is it the negative x axis? I mean by saying x is less than or equal to 0, it does not necessarily mean the negative x axis, it could be any ray going parallel to the x axis (beginning at 0 and going to negative infinity). So which of all this infinite amount of rays does the problem mean? ... Or what am I missing here?

Also, how could I actually start showing that G maps D onto a horizontal strip of width 2pi. I mean I think I can picture it and understand why it is so, but I can't think of a way of doing it using mathematical language.

4. Oct 2, 2008

### Dick

Yes, in this problem they are deleting the negative real axis. You could 'delete' in any number of different ways, as long as you can't circle around 0 without passing through the deleted part. Any ray starting at 0 will work. You seem to be confusing 0 (the point 0+0i) and the y-axis (x=0) when you are talking about 'rays parallel to the x-axis'. There aren't an infinite number that pass through 0. Think about writing the complex numbers in polar form, r*exp(i*theta) and you can choose -pi<theta<pi. The log of that is log(|r|)+i*theta+2*n*pi*i.