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I was trying to solve for $i^i$. I got distracted and did the following:
$e^{i \theta} = \cos(\theta) + i\sin(\theta)$
$e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation:
$\ln(e^{i2\pi}) = \ln(1)$
$i2\pi = 0$
Clearly this is false. What did I do wrong?

Shawhin Layeghi
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    Define what $a^b$ means when $a,b$ are complex first... – Simply Beautiful Art Sep 21 '17 at 22:41
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    the complex exponential is periodic thus not injective, so the logarithm is multivalued. you have $\ln(re^{i\theta})=\ln(r)+i\theta+2ik\pi$ with $k\in\mathbb Z$. – zwim Sep 21 '17 at 22:42
  • The numbers we know as $pi$ and $e$ are just the result of an infinite series of computations. When you use $i$ on this number, you apply it on each computational step individually, rather on the final result of the computation. Since $i$ becomes interchangeably positive and negative, real and imaginary as you multiply it by itself, it means it has a unique effect and is applied in a unique way at each computational step. – Alpha_Pi Sep 21 '17 at 22:52
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    Say, I've got another example:$$1^0=1^2\\therefore0=2\phantom\therefore\~\(-1)^0=(-1)^2\\therefore0=2\phantom\therefore$$ – Simply Beautiful Art Sep 21 '17 at 22:54
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    Since $\cos(2\pi)=\cos(0)$, can we conclude that $2\pi=0$? – Thomas Andrews Sep 21 '17 at 23:52
  • The function $f:\mathbb{C} \to\mathbb{C} $ given by $f(z) =e^{z} $ is not one-one. The function maps both the points $2\pi i$ and $0$ to the value $1$. To give an analogy, if two individuals have same father it does not mean that they are the same individual. – Paramanand Singh Sep 22 '17 at 02:24
  • @user504882: it is difficult to make sense of your comment especially in the context of the current question. – Paramanand Singh Sep 22 '17 at 02:27

3 Answers3

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You cannot conclude that $2i\pi=0$.

If $f$ is an injective function, you can indeed argue that $f(x)=f(y)$ implies $x=y$. However the exponential function is not injective. Indeed, it is periodic with period $2i\pi$: $$ e^{z+2i\pi}=e^z $$ for every complex $z$. The complex logarithm is a wild beast, compared to the standard functions: it cannot be the inverse of the exponential exactly because the exponential is not injective; it's an example of a multivalued function and cannot be used like standard functions.

egreg
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You should ask yourself, if you have understood the definitions of all the relevant objects?

It turns out, that defining $\ln$ is not so easy to do for complex numbers, and that a definition of the $\ln$ does not have the nice properties that they have for real numbers.

For example $$ \ln(e^z) = z $$ does not need to be true for a complex number $z$. (this can be regarded as a mistake in your question)

Just because something is true for real numbers, it doesnt mean that the same thing is also true for complex numbers.

supinf
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You went wrong in assuming $z\to e^z $ was one to one. As $e^{z+2\pi i}=e^z*e^{2\pi i}=e^z (\cos 2\pi +i\sin 2\pi)=e^z*1=e^z$, it clearly is not one-to-one .

Thus we can not assume $e^z=e^w\implies z=w $.

And thus if we define $\ln w=v $ if $e^v=w$, it is not a single value unique function.

In fact $z=a+bi;a,b\implies \mathbb R $ implies $e^z=e^a (\cos b +i\sin b)=e^a (\cos (b+k2\pi)+i\sin (b+k2\pi))=e^{z+2k\pi i} $ for some integer $k $. Thus we can conclude $e^v=e^w $ implies $v=w+2k\pi i$.

So when we define $\ln v=w $ if $e^w=v $ it isn't a single value function. It is a "multivalue function" where $\ln z $ does not just equal one value $u$. It equals an infinite number of values of complex numbers, some multiple of $2i\pi $ apart, of which $u $ is just one example.

So $e^{2\pi i}=e^0=1\implies \ln e^{2\pi i}=\ln 1\implies 2\pi i=0 +2k\pi i$ for some integer $k$. Which it does!

fleablood
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