# Simplifying natural log of complex number

1. Oct 3, 2015

### Sturk200

1. The problem statement, all variables and given/known data
The problem is to sketch lines of constant u and v in the image plane for the function Log[(z+1)/(z-1)].

2. Relevant equations

z=x+iy

3. The attempt at a solution

In order to do this I have to get the expression into u+iv form, so that I can read off and manipulate the u and v aspects of the function. What I can't figure out is how to get this equation into u+iv form.

I know that for Ln[z], I can write Ln[reitheta] = Ln[r] + itheta. Then u=Ln[r] and v=theta. But having that added +1 in the natural log is really throwing me off. Does anyone have any ideas for how to get this into u+iv form? Is there some obvious rule for simplifying the logarithm of a sum that I'm forgetting?

Thanks.

2. Oct 3, 2015

### pasmith

$\ln f(z) = \ln |f(z)| + i\arg(f(z))$.

It might be easier to write $\ln\frac{z + 1}{z-1} = \ln(z+1) - \ln(z - 1)$ first.

3. Oct 3, 2015

### Sturk200

Thanks for your response. I think I'm starting to get it now that I've seen it in general form. Here is what I'm getting:

The modulus of (z+1) is [(x+1)2+y2]1/2, since 1 is a real number and therefore addends itself to the x-component of the modulus.

The argument of z+1 remains unchanged and is equal to theta.

Therefore ln(z+1) = ln[(x+1)^2 + y^2]^1/2 + i(theta).

Is that on the right track?

I'm also wondering how you go about justifying the generalization in the first line of your post. It's clearly true when f(z)=z, but how do you get from there to the general case of taking the modulus and the argument more abstractly?

4. Oct 3, 2015

### pasmith

Are you sure? $\tan (\arg(z+1)) = \frac{y}{x+1}$ which is not in general equal to $\frac yx$.

$z$ in $\ln z = \ln |z| + i\arg(z)$ is just a (non-zero) complex number. You can replace it with any other collection of symbols which represent a (non-zero) complex number and the equation still holds.

5. Oct 3, 2015

### Sturk200

O boy, that was pretty dumb. I think I'm actually getting it now.

So ln(z+1) = ln[(x+1)^2+y^2] 1/2+ i[arctan(y/(x+1))]. Is that right?

That makes so much sense, I don't know why I had a block on it before. Thanks again!