# Thread: Branch Cut question for a complex logarithm function (i think i understand it)

1. ## Branch Cut question for a complex logarithm function (i think i understand it)

Hi there.

I'm new to this whole branch cut concept and was working on the following question. I want to be sure my logic is correct.

The question was...

consider a branch of logz corresponding to the branch cut x = 0, y >= 0 and value log(-1) = -iπ. I must calculate log(-(3^1/2) + i) relative to this branch.

From the little information I have gathered about branch cuts, i have deduced (possibly incorrectly) that The "branch cut"
x = 0, y >= 0 starts the branch at -3π/2 and ends it at π/2 because it must contain a z with it's argument equal to -π

I then went along to get the general
log(-(3^1/2) + i) and found it to be...

ln2 + i(5π/6 +2kπ)

since it must lie on the branch i chose...

ln2 + i(5π/6 - 2π)

= ln2 - 7πi/6

Am I getting this? I'm sorry but I came late to this lecture and had to sit in the back where i couldn't see half the board because of the guy with the fat head in front of me.

2. Originally Posted by douber
Hi there.

I'm new to this whole branch cut concept and was working on the following question. I want to be sure my logic is correct.

The question was...

consider a branch of logz corresponding to the branch cut x = 0, y >= 0 and value log(-1) = -iπ. I must calculate log(-(3^1/2) + i) relative to this branch.

From the little information I have gathered about branch cuts, i have deduced (possibly incorrectly) that The "branch cut" x = 0, y >= 0 starts the branch at -3π/2 and ends it at π/2 because it must contain a z with it's argument equal to -π

I then went along to get the general log(-(3^1/2) + i) and found it to be...

ln2 + i(5π/6 +2kπ)

since it must lie on the branch i chose...

ln2 + i(5π/6 - 2π)

= ln2 - 7πi/6

Am I getting this? I'm sorry but I came late to this lecture and had to sit in the back where i couldn't see half the board because of the guy with the fat head in front of me.
The given branch cut means that the principle argument you need to use lies between -pi and pi.

3. I am not quite sure I understand why it is this, but i can almost see how it works

so if its -pi to pi is that not just the principal value?

ln2 + i(5π/6)

4. Originally Posted by douber
I am not quite sure I understand why it is this, but i can almost see how it works

so if its -pi to pi is that not just the principal value?

ln2 + i(5π/6) Mr F says: This is the correct answer to the question.
..

5. thanks Mr. F i thought i may be on close to the right track. Sadly my book and my notes and my proffessor are all horrible resources for this branch thing.

they all start pulling words like strip and cuts and all that right out of their bottom... without really explaining any of the basics... and the funny thing is... i cant find the basics anywhere. I've only learned what i know by pasting together peices. seems like the most unpopular subject in math.

6. Originally Posted by douber
Hi there.

I'm new to this whole branch cut concept and was working on the following question. I want to be sure my logic is correct.

The question was...

consider a branch of logz corresponding to the branch cut x = 0, y >= 0 and value log(-1) = -iπ. I must calculate log(-(3^1/2) + i) relative to this branch.

From the little information I have gathered about branch cuts, i have deduced (possibly incorrectly) that The "branch cut"
x = 0, y >= 0 starts the branch at -3π/2 and ends it at π/2 because it must contain a z with it's argument equal to -π

I then went along to get the general
log(-(3^1/2) + i) and found it to be...

ln2 + i(5π/6 +2kπ)

since it must lie on the branch i chose...

ln2 + i(5π/6 - 2π)

= ln2 - 7πi/6

Am I getting this? I'm sorry but I came late to this lecture and had to sit in the back where i couldn't see half the board because of the guy with the fat head in front of me.
Based on what he said, this is how I interpret it: I assume you mean the branch-cut is along the positive imaginary axis with $\log(-1)=-\pi i$. Therefore, I would define this branch of logarithm as:

$\text{mylog}(z)=\ln|z|+i\Theta,\quad -3\pi/2<\Theta\leq \pi/2$

Therefore:

$\text{mylog}(-\sqrt{3}+i)=\ln(2)-7\pi i/6$

I don't see how it's $5\pi/6$. Sorry.

7. Originally Posted by shawsend
Based on what he said, this is how I interpret it: I assume you mean the branch-cut is along the positive imaginary axis with $\log(-1)=-\pi i$. Therefore, I would define this branch of logarithm as:

$\text{mylog}(z)=\ln|z|+i\Theta,\quad -3\pi/2<\Theta\leq \pi/2$

Therefore:

$\text{mylog}(-\sqrt{3}+i)=\ln(2)-7\pi i/6$

I don't see how it's $5\pi/6$. Sorry.
You're right. My mistake, I misread the initial post.

8. Originally Posted by shawsend
Based on what he said, this is how I interpret it: I assume you mean the branch-cut is along the positive imaginary axis with $\log(-1)=-\pi i$. Therefore, I would define this branch of logarithm as:

$\text{mylog}(z)=\ln|z|+i\Theta,\quad -3\pi/2<\Theta\leq \pi/2$

Therefore:

$\text{mylog}(-\sqrt{3}+i)=\ln(2)-7\pi i/6$

I don't see how it's $5\pi/6$. Sorry.
Ok so my initial assumptions were correct? yay? hehe