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Euler's Identity - What's the falacy?

We know that \(e^{i\pi} = -1 \). So, \(\ln(-1) = i\pi \). Well, \(\ln(1) = 0\). So, I guess we can write \[\ln(-1) + \ln(-1) = \ln(-1\cdot -1) = \ln(1) = 0\]. But \[\ln(-1) + \ln(-1) = i\pi + i\pi = 2i\pi \]

What's the falacy?

Note by Gabriel Laurentino
3 years, 5 months ago

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the identity you used is applicable for numbers belonging to the set (0,infinity) only

Vivek Kushal - 3 years, 5 months ago

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May be because the properties of ln is applicable only for numbers within its domain..

Kushagra Jaiswal - 3 years, 5 months ago

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I thought the same. Thanks!

Gabriel Laurentino - 3 years, 5 months ago

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