Complex solutions to $1^x = 2$

Question

I saw a YouTube video where the presenter claimed that there were solutions to the equation $1^x = 2.$

To summarize their argument:

Clearly, $1^x = 1$ for all real $x.$ However, there may be complex solutions. $$ \begin{align} &1 = e^{2k\pi i} \\ &1^x = \left(e^{2k\pi i}\right)^x \\ &e^{2kx\pi i} = 2 \end{align} $$ Taking the log of both sides.
$$ \begin{align} 2kx\pi i &= \log 2\\ x & = \frac {\log 2}{2ki} = -\frac {i\log 2}{2k} \end{align} $$ My intuition says that this must be wrong, but I am having a hard time articulating why.
Let's start here: $$ 1^x = \left(e^{2k\pi i}\right)^x $$ This statement is true for integer values of $x,$ but taking $x = \frac 12$ and $k = 1$ we get $1^\frac 12 = -1$

Annother sticking point in my intuition is that $|r^z| = |r|^{\mathrm{Re}(z)}$ for real $r,$ and complex $z.$ And the work above contradicts that.
At the same time, I am fine with the statement $1 = 2^{\frac {2\pi i}{\log 2}},$ which would imply a solution to $2^{\frac 1x} = 1$

I think that the precise answer is that the complex logarithm is a subtle and devious operator. The complex logarithm is multi-valued and as such we must be careful in choosing which branch we are applying. But, I still don't feel like this is squarely hitting the point.

Could someone help me put some rigor and formality around my misgivings? Or, if I am just wrong, clarify where I am wrong.

Answer 1

No, this definitely doesn't work. It's true that

$$1 = e^{2k\pi i}$$

when $k$ is an integer. When you “raise both sides to a power” that is an arbitrary $x$, the condition of $k$ being an integer fails. The general statement is

$$z = e^{a+bi} = e^a\cdot e^{bi}$$

where $a$ and $b$ are real. Then the $e^a$ part is a real number, whose magnitude is the same as that of $z$: $$|e^a| = |z|$$

and the $e^{bi}$ part is a complex number of magnitude $1$.

The only solutions of $e^{a+bi} = 2$ are

$$e^{\log 2 + 2k\pi i} = 2$$

where $e^{\log 2}$ has magnitude $2$, and $e^{2k\pi i}$ has magnitude $1$.

Answer 2

Im going to answer here (despite not really having the background to do so), because I asked the exact same question, and got a different answer (from two people) which concerned me.

Also, assuming me and the OP watched the same video, I believe he hasan't replicated the presenters argument correctly, but I might be missremenbering.

TLDR: The asnwers I got basically said that, while the presenters argument was questionable, "solutions" did exist if we looked at $1^{x}$ as an multivalued complex function, in which case solutions exist for all branches (except the principal one), while the comment above claims no solutions exist at all.

It's also interesting to note wolfram alpha will solve $1 = 2^{\frac{1}{z}}$, but not $1^{z} = 2$.