Understanding Complex Logarithms and Justifying Formulas

  • Thread starter Thread starter Hertz
  • Start date Start date
  • Tags Tags
    Formula
Click For Summary
The discussion focuses on the complexities of understanding and justifying formulas related to complex logarithms. A participant presents the formula ln(-a) = ln(a) + πi and seeks validation for their reasoning, which hinges on Euler's identity. The conversation highlights that complex logarithms are multivalued and that the principal branch is often restricted to avoid ambiguities, leading to different outputs based on term organization. It is emphasized that familiar logarithmic properties may not hold in the complex domain, necessitating careful justification when applying these rules. The dialogue concludes with a recognition of the challenges and nuances involved in working with complex logarithms.
Hertz
Messages
180
Reaction score
8
I've read about complex logarithms and maybe it's just my reading skill, but it all seems to be way too complicated. Anyways, after doing some thinking of my own I've come up with a few formulas, one of which is ln(-a) = ln(a) + πi; however, my reasoning behind such formulas seems way too basic, so I'm doubting myself. Could someone please check my work and let me know if what I'm doing is ok?

Here's the derivation:
ln(-a) = ln(-a)
ln(-a) = ln(-1a)
ln(-a) = ln(-1) + ln(a) -- Properties of Logs
ln(-a) = πi + ln(a) -- Euler's Identity

Euler's identity states e^(iπ) = -1; therefore, ln(-1) = iπ.
 
Mathematics news on Phys.org
So long as you restrict a to being positive and real, that formula should be valid. Well, up to a multiple of 2πi anyway.
 
Char. Limit said:
So long as you restrict a to being positive and real, that formula should be valid. Well, up to a multiple of 2πi anyway.

Wait a minute, that's an excellent point.. Why does the output of the function change depending on how you organized the terms?

ln(-a) = πi + ln(a)
If we attempt to let a = -1
ln(1) = πi + ln(-1)
ln(1) = 2πi

However, if we write ln(-1) = πi + ln(a) as
ln(a) = ln(-a) - πi
Now if we let a = 1 to solve for ln(1)
ln(1) = ln(-1) - πi
ln(1) = πi - πi
ln(1) = 0

Is 2πi the same as zero? Looking at Euler's Identity I've never seen the π as evidence of a cycle, I've always just thought of it as a constant. Interesting... That's something to think about..

e-
The formula is intended to work for all a, a != 0. Why must a be limited to positive values?
 
The complex logarithm is a tricky animal. First of all, it is multivalued. This can easily be seen by

e^0=1=e^{2\pi i}

So the logarithm of 1 should be both 0 and 2\pi i (and other values).

A formula for the complex logarithms can be given by

Log(z)=log|z| + i arg(z) + 2\pi i k

for all k\in \mathbb{Z} (that is: each k gives a good value for the logarithm).

The arg(z) is the angle that z makes with the positive x-axis in radians.

For example, if z is a positive real, then it makes a 0 angle with the positive x axis, and thus

Log(z)=log(z)+2\pi i k

if k=0, then this corresponds with the ordinary logarithm.

If z is a negative real, then it makes a \pi (or -\pi, doesn't matter) angle with the positive x-axis. And thus

Log(z)=log(-z)+\pi i + 2\pi i k

If k=0, then we get the formula you discovered.

Mathematicians don't like multi-valued functions, that's why they restrict the logarithm to only take on one value. That is: we always take k=0 in the previous formula. This is called the principal branch of the logarithm. This is a very arbitrary choice and other choice can be made. But it has as benefit that the complex logarithm coincides with the real logarithm for positive real numbers.

Also, complex logarithms are dangerous because many familiar laws do not hold anymore. For example, Log(zw)=Log(z)+Log(w) is not valid anymore (certainly not for the principal branch logarithm).
 
micromass said:
The complex logarithm is a tricky animal. First of all, it is multivalued. This can easily be seen by

e^0=1=e^{2\pi i}

So the logarithm of 1 should be both 0 and 2\pi i (and other values).

A formula for the complex logarithms can be given by

Log(z)=log|z| + i arg(z) + 2\pi i k

for all k\in \mathbb{Z} (that is: each k gives a good value for the logarithm).

The arg(z) is the angle that z makes with the positive x-axis in radians.

For example, if z is a positive real, then it makes a 0 angle with the positive x axis, and thus

Log(z)=log(z)+2\pi i k

if k=0, then this corresponds with the ordinary logarithm.

If z is a negative real, then it makes a \pi (or -\pi, doesn't matter) angle with the positive x-axis. And thus

Log(z)=log(-z)+\pi i + 2\pi i k

If k=0, then we get the formula you discovered.

Mathematicians don't like multi-valued functions, that's why they restrict the logarithm to only take on one value. That is: we always take k=0 in the previous formula. This is called the principal branch of the logarithm. This is a very arbitrary choice and other choice can be made. But it has as benefit that the complex logarithm coincides with the real logarithm for positive real numbers.

Also, complex logarithms are dangerous because many familiar laws do not hold anymore. For example, Log(zw)=Log(z)+Log(w) is not valid anymore (certainly not for the principal branch logarithm).

Hmm, so I should be sure when using properties of logs that I don't perform operations on imaginary logs unless I can justify the formula in the imaginary plane first?

I've seen this formula before but it's one of those ones where I take a look at it and shy away, but really it's not all that difficult. This post will surely give me a lot to think about. Thank you

(I think what I'm going to do next is try to generalize the familiar log formulas.)

P.s. The log above is considered to be the natural log right? Or is that formula a generalization for all bases?

e-
Two questions:
Is this operation ok?
ln(a) = ln(-1) + ln(-a)?
This is a basic rule of logarithms, but does the ln(-1) make the use of the rule questionable?

Do you have a proof of that formula? I'd like to see how it was derived.
 
Last edited:
Hertz said:
Hmm, so I should be sure when using properties of logs that I don't perform operations on imaginary logs unless I can justify the formula in the imaginary plane first?

Yes. I think most properties will hold for the multivalued log, but not for the principal branch. But it needs to be proven first.

I've seen this formula before but it's one of those ones where I take a look at it and shy away, but really it's not all that difficult. This post will surely give me a lot to think about. Thank you

It's really not hard. Take a complex number w. Denote

Log(w)=z

this means by definition that

e^z=w

Write z=z_1+iz_2. We can write w in its polar form as |w|(\cos\theta +i\sin\theta) where \theta is the arg of w. Thus we have

e^{z_1+iz_2}=|w|(\cos\theta+i\sin\theta)

By Eulers formula, we have

e^{z_1+iz_2}=e^{z_1}e^{iz_2}=e^{z_1}(\cos z_2 + i\sin z_2)

So we got the equation

e^{z_1}(\cos z_2 + i\sin z_2)=|w|(\cos\theta +i\sin\theta)

We deduce from this that e^{z_1}=|w| and thus z_1=log|w|. Also, we have that

\cos z_2 = \cos\theta~\text{and}~\sin z_2=\sin\theta

This implies that z_2=\theta+k2\pi for a certain k. Thus

Log(w)=log|w|+i(\theta+k2\pi)

which is the formula we wanted.

P.s. The log above is considered to be the natural log right? Or is that formula a generalization for all bases?

That is a formula for the natural log. I'm sure there are generalizations for other bases. But for some reason, other bases don't really show up much in complex analysis.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 3 ·
Replies
3
Views
540
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 3 ·
Replies
3
Views
4K