Contradiction in Euler’s Identity? $2i\pi = 0$

Question

Many of us know Euler’s famous identity:

$e^{i\pi} + 1 = 0$.

But if we add -1 (subtract 1) to both sides we get:

$e^{i\pi} = -1$

then natural log of both sides and:

$i\pi = \ln(-1)$

Next we multiply both sides by 2:

$2i\pi = 2\ln(-1)$

Which by basic logarithm rules is equal to:

$2i\pi = \ln({-1}^2) = \ln(1) = 0$

so

$2i\pi = 0$

which can’t be true as either 2, i or $\pi$ would have to equal 0. My best guess is that logarithms just plainly aren’t defined for negative values, I just assumed this was just true logarithms bounded under the real numbers, though I might be wrong. If someone could help it would be great.

Thanks.

Natural logarithm is for positive numbers, not complex numbers. Note that $e^{0}=e^{2\pi i}$ but $0 \neq 2\pi i$. — Kavi Rama Murthy, Mar 13 '19 at 06:42
Yeah, and as said below, it is defined, but differently. Thanks. — , Mar 13 '19 at 06:43
What i mean is that the multi-valued logarithm in the complex plane is not called natural logarithm and properties of natural logarithm don't hold for it. — Kavi Rama Murthy, Mar 13 '19 at 06:46

Fabian · Accepted Answer · 2019-03-13T06:43:32.813

3

The (complex) logarithm is a multivalued function. The solution to $$ e^{z} = w$$ is given by $$ z= \log w + 2\pi i n$$ where $n$ is an arbitrary integer, $n=\dots, -2,-1,0,1,2,\dots$.

In your case, you will get $$ 2\pi i = 0 + 2\pi in$$ which is clearly no contradiction (as $n=1$ makes this an equality).

edited Mar 13 '19 at 06:43

answered Mar 13 '19 at 06:42

Fabian

24,230

Could you please edit this so it gives a definitively answer... thanks – Mar 13 '19 at 06:43
Thanks. That is better. – Mar 13 '19 at 06:44
@L.McDonald If you feel it is a definitive answer then mark it with a $\color{green}\checkmark$ – gen-ℤ ready to perish Mar 13 '19 at 07:36
I know, but it was late at night for me, and I didn’t have time for the 10 minute break. – Mar 14 '19 at 04:49

Contradiction in Euler’s Identity? $2i\pi = 0$

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