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I don't really understand the step by step procedure on how to calculate this? That $-$ confuses me, because I don't understand how there can be an $\ln$ of a negative number. Can someone please explain? Thanks

Leucippus
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ivana14
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  • Are you sure you have this question correct? You cannot take logarithms of negative numbers. Where did you find this question? – Ananth Kamath May 12 '17 at 06:52
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    $\ln(-e)=\ln(-1)+1$. Now use $e^{i (2k+1)\pi} = -1$ – jonsno May 12 '17 at 06:52
  • $\ln$ is only defined for positive numbers. – 5xum May 12 '17 at 06:52
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    That's correct, in the real numbers you can't take the logarithm of a negative number. It's likely that what you think the question is asking is not what it's really asking. Why don't you provide more context? – Erick Wong May 12 '17 at 06:52
  • @ErickWong I don't have any more context, I have just been given this problem to solve. An online calculator gives me the answer $1 + i \pi$, but I am not sure how. – ivana14 May 12 '17 at 06:58
  • There isn't any real logarithm of a negative number. And in the complex numbers the definition of a logarithm is derived in a different manner. In a nutshell, a complex number can be expressed as $e^{a+ib}=e^a (\cos b + i \sin b) $ in this case $e^{a+ib}=e^1 (-1+i0) $ so $a=1$ and $b=(2k+1)\pi $ and $\ln -e = 1+i (2k+1)\pi $ for any integer $k $. – fleablood May 12 '17 at 07:02

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It looks like you are referring to the complex logarithmic function which is defined as:

$\ln(z) = \ln(r) + i(\theta+2\pi k)$, where $z = re^{i\theta}, k\in \mathbb Z$.

Applying that here, since $-e = e\times e^{i\pi}$, then $\ln(-e) = \ln(e)+ i(\pi+ 2\pi k) = 1 + i(\pi+ 2\pi k)$.

Leucippus
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erdoswiles
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Indeed, from Euler's equation and more specifically Euler's identity you can define a logarithm of a negative number.

If you had $\ln (-1)$, all you would have to do is add it to $\ln (n)$ to get $\ln (-n)$. The axioms defined for logs remain the same.

Euler's equation: $e^{i\pi}=-1 \implies \ln({-1})=i\pi$

I know there's a periodicity of $2\pi$ that I haven't accounted for yet but I think that's a fair enough point made.

I assume you're not that familiar with complex numbers so like any good fellow, I wish you good tidings on the yellow brick road you will trod in complex numbers.

What they are: https://www.youtube.com/playlist?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF

Why $e$ shows up: https://youtu.be/mvmuCPvRoWQ

Nεo Pλατo
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