ln⁡(−1)\ln(-1)

let z=ln⁡(−1)ez=−1ei(zi)=−1cos⁡(zi)+isin⁡(zi)=−1ℜ(cos⁡(zi)+isin⁡(zi))=ℜ(−1)cos⁡(zi)=−1zi=(2n−1)πz=(2n−1)πi\begin{aligned} \text{let}\space z&=\ln(-1)\\ e^z&=-1\\ e^{i\left(z\over i\right)}&=-1\\ \cos\left(z\over i\right)+ i\sin\left(z\over i\right)&=-1\\ \Re\left(\cos\left(z\over i\right)+ i\sin\left(z\over i\right)\right)&=\Re\left(-1\right)\\ \cos\left(z\over i\right)&=-1\\ z\over i&=(2n-1)π\\ z&=(2n-1)πi \end{aligned}

∴ln⁡(−1)=(2n−1)πi\therefore \boxed{\ln(-1)=(2n-1)πi}

#Algebra

Note by Gandoff Tan
1 year, 9 months ago

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Comments

Knowing that: eia=cos⁡(a)+isin⁡(a)e^{ia}=\cos(a) + i\sin(a)

ei(2k+1)π=cos⁡((2k+1)π)+isin⁡((2k+1)π)e^{i(2k+1)\pi} = \cos((2k+1)\pi) + i\sin((2k+1)\pi)

Where kk is any non-negative integer. This leads to:

ei(2k+1)π=cos((2k+1)π)=−1e^{i(2k+1)\pi} = cos((2k+1)\pi) = -1

Which means, taking natural log on both sides:

ln⁡(ei(2k+1)π)=ln⁡(−1)\ln(e^{i(2k+1)\pi}) = \ln(-1)

i(2k+1)π=ln⁡(−1)\boxed{i(2k+1)\pi = \ln(-1)}

This leads to the conclusion that ln⁡(−1)\ln(-1) is not a unique complex number. The boxed equation is a generalised result.

Karan Chatrath - 1 year, 9 months ago

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I even posted a problem on this, which should be pretty easy to tackle, given the above analysis.

https://brilliant.org/problems/natural-log-of-a-negative-number/?ref_id=1573413

Karan Chatrath - 1 year, 9 months ago
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