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Simplifying natural log of complex number

  • Thread starter Sturk200
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  • #1
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Homework Statement


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

Homework Equations



z=x+iy

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.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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[itex]\ln f(z) = \ln |f(z)| + i\arg(f(z))[/itex].

It might be easier to write [itex]\ln\frac{z + 1}{z-1} = \ln(z+1) - \ln(z - 1)[/itex] first.
 
  • #3
168
14
[itex]\ln f(z) = \ln |f(z)| + i\arg(f(z))[/itex].

It might be easier to write [itex]\ln\frac{z + 1}{z-1} = \ln(z+1) - \ln(z - 1)[/itex] first.
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
pasmith
Homework Helper
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410
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.
Are you sure? [itex]\tan (\arg(z+1)) = \frac{y}{x+1}[/itex] which is not in general equal to [itex]\frac yx[/itex].

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?
[itex]z[/itex] in [itex]\ln z = \ln |z| + i\arg(z)[/itex] 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
168
14
Are you sure? tan(arg(z+1))=yx+1\tan (\arg(z+1)) = \frac{y}{x+1} which is not in general equal to yx\frac yx.
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?

zz in lnz=ln|z|+iarg(z)\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.
That makes so much sense, I don't know why I had a block on it before. Thanks again!
 

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