MHB The Controversy Surrounding the Definition of Logarithm for Complex Numbers

[Suvadip]

In the context of complex number, how to prove that

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

 

[ThePerfectHacker]

In the context of complex number, how to prove that

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

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.

 

[alyafey22]

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 $Arg(z) \in (-\pi , \pi ]$.

 

[alyafey22]

In the context of complex number, how to prove that

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

Generally we have the following

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

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

The proof is not difficult especially when we prove that

$$arg(z_1 z_2)= arg(z_1) +arg(z_2)$$

But remember that

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

Can you give counter examples ?

 

[polygamma]

$ \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} $

 

[chisigma]

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 $\chi$ $\sigma$

 

[alyafey22]

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

[math]Log(z) = \ln |z|+i Arg(z) [/math] 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 .