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- Thread starter suvadip
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- Jan 26, 2012

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It depends how you depend a complex logarithm. The standard definition is, for $z\not = 0$,In the context of complex number, how to prove that

1. \(\displaystyle log i +log(-1+i) \neq log i(-1+i)\)

2. \(\displaystyle log i^2 =2log i\)

$$ \log z = \log |z| + i\arg z$$

Where $\arg z$ angle in interval $(-\pi,\pi]$.

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

- Jan 17, 2013

- 1,667

You mean the principle logarithm ? It is actually customary to denote that with capital A for the argument hence $Arg(z) \in (-\pi , \pi ]$.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.

- Jan 17, 2013

- 1,667

Generally we have the followingIn the context of complex number, how to prove that

1. \(\displaystyle log i +log(-1+i) \neq log i(-1+i)\)

2. \(\displaystyle log i^2 =2log i\)

\(\displaystyle \log(z_1 z_2) = \log(z_1)+\log(z_2) \) 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 ?

- Jan 31, 2012

- 253

$\text{Log}(z_{1}^{n}) = n \text{Log}(z_{1})$ iff $ -\frac{\pi}{n} < \text{Arg}(z_{1}) \le \frac{\pi}{n} $

- Feb 13, 2012

- 1,704

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 wich 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

$\chi$ $\sigma$

- Jan 17, 2013

- 1,667

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

$\chi$ $\sigma$

\(\displaystyle Log(z) = \ln |z|+i Arg(z) \) 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 .