Natural Logarithm contradiction?

Question

Please help with this contradiction.

• $e^{i\pi}=-1$
• $e^{i2\pi}=1$
• $\ln(1)=i2\pi$
• $0=i2\pi$

Answer 1

The complex logarithm is multi-valued.

$$e^{i2\pi}=1\leftrightarrow i2\pi=\log1+i2k\pi.$$

Answer 2

Bill says "I am the son of Ted Clark". John says "I am the son of Ted Clark". DNA testing shows they are both correct.

Therefore: Bill = John.

Where's the contradiction?

...

Son(Ted CLark) = John.

Ted Clark = Father(Bill)

so

Son(Ted Clark) = Son(Father(Bill)) = Bill

But Son(Ted Clark) = John.

====

With complex numbers $e^z$ is not injective (one to one) so it is possible for $e^z = e^w$ but $w \ne z$. There is nothing at all contradictory about that at all.

So an inverse function of $e^x$ is not definable as single value/unlimited range function.

A similar "paradox" would be:

So $\arcsin ( \frac {\sqrt 2}2) = \arcsin (\frac {\sqrt 2} 2)$

$\arcsin (\sin \frac {3\pi}4) = \arcsin (\sin \frac {\pi}4)$

$\frac {3\pi} 4 = \frac {\pi} 4$.

In the case of $\arcsin $ the definition is not "$\arcsin x$ is the $\theta$ so that $\sin \theta = x$". The definition is "$\arcsin x$ is the $\theta$ WITHIN $-\frac {\pi}2 < \theta \le \frac {\pi} 2$ so that $\sin \theta = x$.

So $\arcsin (\sin \theta) = \theta$ and $\arcsin (\sin \frac {3\pi} 4) = \frac {3\pi}4$ is simply not true at all.

(Another example of this is the definition of $\sqrt{}$. The definition is not $\sqrt{x} = y$ so that $y^2 = x$. It is $\sqrt{x} = y$ so that $y \ge 0; y^2 = x$.... assuming we are talking about real numbers. SO $\sqrt {(-2)^2} \ne -2$.)

Now... We do something slightly different with natural log.

If $e^z = 1$ then there are an infinite set of values of complex numbers that $z$ could be. We could have $z = 0$ but we could also have $z = 2\pi i$ or $z \in \{2k\pi i|k\in \mathbb Z\}$.

We choose to define $\ln$ not be one representative value chosen from the set (which is what we do four $\sqrt{}$ or $\arcsin$) but to define $\ln$ as a "multi-valued" function. That is a function with more than just a single output for a single input. Ln $1$ is any of the values of $z \in \{2k\pi i|k\in \mathbb Z\}$.