The Controversy Surrounding the Definition of Logarithm for Complex Numbers

In summary, the conversation discusses the definition of complex logarithm, which is given by $\log z = \log |z| + i\arg z$. However, some rules for logarithms may not work for complex logarithms. The conversation also discusses the incorrectness of the standard definition and provides a counterexample. The conversation concludes with a discussion about the branch cut for $z>0$ and the different values obtained when approaching the integral from above and below.
  • #1
Suvadip
74
0
In the context of complex number, how to prove that

1. \(\displaystyle log i +log(-1+i) \neq log i(-1+i)\)
2. \(\displaystyle log i^2 =2log i\)
 
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  • #2
suvadip said:
In the context of complex number, how to prove that

1. \(\displaystyle log i +log(-1+i) \neq log i(-1+i)\)
2. \(\displaystyle log i^2 =2log i\)

It depends how you depend a complex logarithm. The standard definition is, for $z\not = 0$,
$$ \log z = \log |z| + i\arg z$$
Where $\arg z$ angle in interval $(-\pi,\pi]$.

Note, many rules for logarithms you are used to need not work for complex logarithms.
 
  • #3
ThePerfectHacker said:
It depends how you depend a complex logarithm. The standard definition is, for $z\not = 0$,
$$ \log z = \log |z| + i\arg z$$
Where $\arg z$ angle in interval $(-\pi,\pi]$.

Note, many rules for logarithms you are used to need not work for complex logarithms.

You mean the principle logarithm ? It is actually customary to denote that with capital A for the argument hence $Arg(z) \in (-\pi , \pi ]$.
 
  • #4
suvadip said:
In the context of complex number, how to prove that

1. \(\displaystyle log i +log(-1+i) \neq log i(-1+i)\)
2. \(\displaystyle log i^2 =2log i\)

Generally we have the following

\(\displaystyle \log(z_1 z_2) = \log(z_1)+\log(z_2) \) where $\log$ defines the multiple valued function

\(\displaystyle \log(z) = \ln|z|+i arg(z) \)

The proof is not difficult especially when we prove that

\(\displaystyle arg(z_1 z_2)= arg(z_1) +arg(z_2)\)

But remember that

\(\displaystyle Arg(z_1 z_2) \neq Arg(z_1) +Arg(z_2) \)

Can you give counter examples ?
 
  • #5
$ \text{Log} (z_{1}z_{2}) = \text{Log}(z_{2}) + \text{Log}(z_{2})$ iff $ - \pi < \text{Arg}(z_{1}) + \text{Arg} (z_{2}) \le \pi$

$\text{Log}(z_{1}^{n}) = n \text{Log}(z_{1})$ iff $ -\frac{\pi}{n} < \text{Arg}(z_{1}) \le \frac{\pi}{n} $
 
  • #6
The so called 'standard definition' of the logarithm of a complex variable z is, in my opinion of course, wrong and the reason of that is explained in the following example... http://mathhelpboards.com/calculus-10/improper-integral-involving-ln-6103.html#post28032

... where the application of such a definition conducts to an erroneous computation of a definite integral which is solvable with elementary method. I realize however that that is a 'delicate' question and it must be discussed 'with calm and reason'... Kind regards $\chi$ $\sigma$
 
  • #7
chisigma said:
The so called 'standard definition' of the logarithm of a complex variable z is, in my opinion of course, wrong and the reason of that is explained in the following example... http://mathhelpboards.com/calculus-10/improper-integral-involving-ln-6103.html#post28032

... where the application of such a definition conducts to an erroneous computation of a definite integral which is solvable with elementary method. I realize however that that is a 'delicate' question and it must be discussed 'with calm and reason'... Kind regards $\chi$ $\sigma$

I don't understand how is that definition questionable. If we use the branch cut for $z>0$ then having the definition

\(\displaystyle Log(z) = \ln |z|+i Arg(z) \) where $z \in (0,2\pi ] $

Then approaching the integral from above gives $Log(z) = \ln (x) $ and $Log(z) = \ln (x) +2\pi i $ when approaching it from below .
 

FAQ: The Controversy Surrounding the Definition of Logarithm for Complex Numbers

What is a logarithm of a complex number?

A logarithm of a complex number is a mathematical operation that gives the exponent of a given base number that results in the complex number. It is the inverse operation of exponentiation.

How do you calculate the logarithm of a complex number?

To calculate the logarithm of a complex number, you can use the formula: logb(z) = ln(z) / ln(b) where z is the complex number and b is the base. Alternatively, you can use a calculator or a computer program to compute the logarithm.

What are the properties of the logarithm of a complex number?

The logarithm of a complex number has the same properties as the logarithm of a real number. These properties include the product rule, quotient rule, power rule, and change of base rule.

Can the logarithm of a complex number be negative?

Yes, the logarithm of a complex number can be negative. This happens when the absolute value of the complex number is greater than 1 and the base is a fraction between 0 and 1.

What are the applications of the logarithm of a complex number in science?

The logarithm of a complex number has various applications in science, including signal processing, electrical engineering, and quantum mechanics. It is also commonly used in finance and economics for calculating compound interest and growth rates.

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