I ##1^x =2## has complex solutions?

[Arjan82]

TL;DR Summary

There's a Youtube video claiming $1^x=2$ has solutions $x= \frac{-i \ln(2)}{2\pi n}$ with $n \in \mathbb{Z}$ and $n \neq 0$. Is this solid?

The Youtuber 'Blackpenredpen' claims that $1^x=2$ has solutions $x= \frac{-i \ln(2)}{2\pi n}$ with $n \in \mathbb{Z}$ and $n \neq 0$. Somone else in a forum claims that because $1^x$ is not injective there are more solution branches and this solution mixes these branches somehow. Who is right?

Also, 'Blackpenredpen' does not show that $1^{ \frac{-i \ln(2)}{2\pi n}}$ is indeed equal to 2. He kind of waved it away with 'something something infinite amount of solutions but one of them is 2'. Is this correct?

 

[Hill]

Let's see.
$1=e^{2\pi i}$.
Take for example $n=1, x=\frac{-i \ln(2)}{2\pi}$.
We get
$1^x=(e^{2\pi i})^{\frac{-i \ln(2)}{2\pi}}=e^{(2\pi i)(\frac{-i \ln(2)}{2\pi})}=e^{ln(2)}=2$.
Looks right.

(Sorry, I don't watch videos.)

 

[fresh_42]

And if $n>1$?

I tend to agree with WA which sees it as an undefined statement.

Before I start any discussion, I want to know how $1^x$ is defined. You can do a lot of nasty tricks with complex numbers that are mathematically just wrong, e.g.
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

(Sorry, I don't watch videos, either.)

 

[pines-demon]

Logarithm has many solutions if you do not take the principal branch.

$$\log(1)=0, 2 \pi i, 4\pi i,....$$
and so on. This video plays on that

 

[Hill]

For $n=2$ I'd take $1=e^{4 \pi i}$, etc.
I also

want to know how $1^x$ is defined.

I remember the general definition, $z^w=e^{w log(z)}$.

 

[pines-demon]

For those that did not watch the video, here it the derivation. Start with
$$1^x=2$$
Take the logarithm
$$\log(1^x)=\log2$$
Here is heavy lifting part:
$$x=\frac{\log2}{\log1}$$
Then wait that's dividing by zero unless:
$$x=\frac{\log2}{\log(e^{i(0+2\pi n)})}=\frac{-i \log2}{2 \pi n},\forall n\in\mathbb{Z}/0$$

 

[fresh_42]

Let's see. I made the mistake of assuming that $x$ is real in my previous posts, sorry.

Say $x=a+ib.$
\begin{align*}
1=e^{2\pi i}&\Longrightarrow 1^x=1^a \cdot 1^{ib}=e^{-2b\pi }=2\\
& \Longrightarrow b = n\cdot i -\dfrac{\log(2)}{2\pi}\;(n\in \mathbb{Z})
\end{align*}
$a$ can obviously be any real number so $x=a -\dfrac{\log(2)}{2\pi}\cdot i$ if I didn't make a mistake again.

 

[WWGD]

Well, within a branch logz of the log, which is a local inverse, $z^w=e^{w logz}$. Branches are( can be seen as) vertical strips of width $2\pi^{-}$, or $[z_0, z_0+ 2\pi)$. Within any of these, the log is in bijection with $\mathbb C-{0}$.

Then , within that branch:
$1^x=e^{x log(1)}=e^{xlog(ln|1|+iarg(1))}$. Then it all depends on what value of arg is assigned at $z=1$. Which ,in this setup, would be within $[ik2\pi, i(k+1)2\pi)$. We then just need an interval $[iy, iy+2\pi)$ containing the value $iln2$

Or, within a slightly different setup, we use
$re^{i\theta}=r(cos(\theta)+isin(\theta))$
So it seems within a value of $\theta=cos^{-1}(ln2)$.

Maybe you can argue that $1^z:=e^{zln(1)}$, is Analytic and thus we can use the mean value to show it assumes the value $2$.