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When I do this contour integral, facing a extreme predicament, I have some difference about the residue part.

Here is my step:

To make it simple, I want to know the extra form of $\log^2 (-1+i)-\log^2 (-1-i)$.

Then it is apparently a form of differences of squares, so I factorize it and get $$[\log (-1+i)+\log (-1-i) ][\log (-1+i)-\log (-1-i) ] = \log ((-1+i) (-1-i)) \log (\frac{-1+i}{-1-i}) = \log 2 \log -i = -\frac{\pi}{2} log 2$$.

But the correct answer gives:

$$\log^2 (-1+i) = \left(\frac{1}{4}\log^2 2 - \frac{9}{16} \pi^2 + \frac{3}{4}\log 2 \times i\pi\right)\tag{1}$$ and $$\log^2 (-1-i) =\left(\frac{1}{4}\log^2 2 - \frac{25}{16} \pi^2 + \frac{5}{4}\log 2 \times i\pi\right)\tag{2}$$ From $(1)-(2)$, $\log^2 (-1+i) -\log^2 (-1-i) = \left(\pi^2 - \frac{1}{2}i\pi \log 2\right)$

I use wolframalpha and its result is mine but obviously, mine is not correct because with that I continued my calculation finally got a different answer.

Can someone help me? I really struggle with this different. Any help, I will thank.

Here $log^2(x) \text{ equals } log(x) \times \ log(x)$

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Zau
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    Usually $\log^2(x)$ means $\log(\log(x))$. Do you use a different notation? – Kenny Lau Jul 30 '16 at 07:06
  • @KennyLau Yes $log^2(x) = log(x) \times \ log(x)$. I think log(log(x)), $log \circ log(x)$ supposes to be. – Zau Jul 30 '16 at 07:07
  • Complex $\log$ is a multi-valued function. If you use another value, you can get $\log^2(-1-i)=\left(\dfrac14\log^22-\dfrac{9}{16}\pi^2-\dfrac34\log2\times i\pi\right)$, which should produce your answer. – Kenny Lau Jul 30 '16 at 07:13
  • Do you have any idea which branch you should use for your contour integral? – Kenny Lau Jul 30 '16 at 07:14
  • My answer to$ \log^2 (-1+i)-\log^2 (-1-i)$ is $-\frac{\pi}{2} log 2$ – Zau Jul 30 '16 at 07:15
  • I know, you don't need to repeat that. I only stated the second term. The first term is the same. – Kenny Lau Jul 30 '16 at 07:15
  • @KennyLau what is branch? I once heard a term called branch cut. Is that it? – Zau Jul 30 '16 at 07:16
  • Do you understand that complex $\log$ is multi-valued? For example, $\log(e^{i\pi})$ can be $i\pi$ or $3i\pi$ or $5i\pi$. – Kenny Lau Jul 30 '16 at 07:16
  • @KennyLau but the key issue is that I use my difference of square method. It comes up with a different result and wolframalpha approve it. – Zau Jul 30 '16 at 07:17
  • I say again. If you use another value of $\log$, you can come up with $\log^2(-1-i)=\left(\dfrac14\log^22-\dfrac{9}{16}\pi^2-\dfrac34\log2\times i\pi\right)$. – Kenny Lau Jul 30 '16 at 07:19
  • If you use this and $(1)$, you will get your answer. – Kenny Lau Jul 30 '16 at 07:19
  • @KennyLau Yes of cause otherwise I will not say that $\log -i = \frac{-\pi}{2} $ – Zau Jul 30 '16 at 07:19
  • So the point is that the "correct answer" used another value of $\log^2(-1-i)$. Do you understand what I am saying? – Kenny Lau Jul 30 '16 at 07:21
  • But I have a question that since log is multi-valued, why $\log^2(-1-i)= (\log(2) - 3/4 \pi i)^2$ rather than $\log^2(-1-i)= (\log(2) - 3/4 \pi i + 2\pi * i)^2$ ? (here star is multiplication) – Zau Jul 30 '16 at 07:22
  • Both should be $\frac34i\pi$ instead of $\frac34\pi$. – Kenny Lau Jul 30 '16 at 07:23
  • And the reason is that you can only use one value for the contour integral. And I'm asking you which value you should use. – Kenny Lau Jul 30 '16 at 07:23
  • Which means, more context on the contour integral please. In which step do you encounter this problem? – Kenny Lau Jul 30 '16 at 07:26
  • @kennyLau When I try to sum the residue of poles. – Zau Jul 30 '16 at 07:28
  • Can you show us the integral that generated the poles? – Kenny Lau Jul 30 '16 at 07:31
  • @kennyLau that's why my first statement there. – Zau Jul 30 '16 at 07:32
  • Can you see the sentence "As we actually do the computation of these residues we need to be careful to use the same branch of the logarithm as in the integral." in the answer? – Kenny Lau Jul 30 '16 at 07:33
  • Can you see the sentence "We use the branch of the logarithm with the cut along the positive real axis and returning an argument from zero to 2π."? – Kenny Lau Jul 30 '16 at 07:37
  • @kennyLau sorry I have not. Thanks a lot. I finally come up with correct answer with my original method and I have one question: what is branch cut for logarithm function. – Zau Jul 30 '16 at 07:40
  • Since $\log$ is multi-valued, we need to use one branch of it, which is another way of saying we need to make it single-valued by choosing one value for every input. – Kenny Lau Jul 30 '16 at 07:41
  • Yes so $\mathbb R \times {u| u = ki, 0 \leq k \leq 2 \pi }$ is a codomain for logarithm function output single value? – Zau Jul 30 '16 at 07:43
  • Yes, that is correct. – Kenny Lau Jul 30 '16 at 07:44
  • @kennyLau it include all the complex number whose imaginary part is within $[0,2\pi]$ – Zau Jul 30 '16 at 07:45

1 Answers1

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$\quad\log^2(-1-i)$

$=[(\log(-1-i))]^2$

$=[(\log(\sqrt2e^{5i\pi/4})]^2$

$=[\frac12\log(2)+5i\pi/4 \ +\ 2ni\pi]^2$

Because complex $\log$ is multi-valued.

If we take $n=0$, we would get $\log^2 (-1-i) =\left(\frac{1}{4}\log^2 2 - \frac{25}{16} \pi^2 + \frac{5}{4}\log 2 \times i\pi\right)$, same as the "correct answer" you referenced.

If we take $n=-1$, we would get $\log^2 (-1-i) =\left(\frac{1}{4}\log^2 2 - \frac{9}{16} \pi^2 - \frac{3}{4}\log 2 \times i\pi\right)$, which would produce your answer.

So, which one should we take?

Notice that in the answer you linked to, it is said that "[w]e use the branch of the logarithm with the cut along the positive real axis and returning an argument from zero to $2π$."

Which means, we should make the complex part of the logarithm between $0$ (inclusive) and $2\pi$ (exclusive).

Which means, we should take $n=0$ instead of $n=-1$.

Kenny Lau
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