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?

#Euler'sFormula(ComplexNumbers) #MathematicalFallacies #Naturallogarithm

Easy Math Editor

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Comments

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

vivek kushal - 7 years ago

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

Kushagra Jaiswal - 7 years ago

I thought the same. Thanks!

Gabriel Laurentino - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$