multiplying two multi-valued quantities We know that $\sqrt {-1}=\{i,-i\}$, then $\sqrt {-1} \sqrt {-1}=\{i,-i\}\{i,-i\}=\{1,-1\}$ or $\sqrt {-1} \sqrt {-1}=\{i^2, (-i)^2\}=\{-1\}$?
 A: One way make sure no funny business is going on let's use Euler's formula:
$$e^{i \pi } +1 =0$$
so $-1 = e^{i \pi }$, but remember that's not the only way to express $-1$. Indeed $\pi$ radians around the unit circle is the same as $3 \pi$ or $5 \pi$ radians etc $\cdots$. So the correct way to express $-1$ is: $$-1 = e^{i(2k+1)}$$ where $k \in \mathbb Z$.
Now that's covered, let's look at
\begin{align}
\sqrt {-1} \sqrt {-1} &= \sqrt {e^{i(2k+1)}} \sqrt {e^{i(2k+1)}} \\
&={e^{i(2k+1)}}^{\frac 12} {e^{i(2k+1)}}^{\frac 12} \\
&=e^{i\frac{2k+1}{2}} e^{i\frac{2k+1}{2}} \ \ , \text {using properties of exponents} \\
&=e^{i(\frac{2k+1}{2} + \frac{2k+1}{2})}, \text {using properties of $e$} \\
&=e^{i(2k+1)} \\
&= -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,\text {by definition}
\end{align}
A: Following Omar's advice, $\sqrt{-1}=e^{i(2k+1)\pi/2}$ for $k\in\mathbb{Z}$, we get
$\sqrt{-1}\sqrt{-1}=e^{i(2m+1)\pi/2} e^{i(2n+1)\pi/2}$,
and not $= e^{i(2k+1)\pi/2} e^{i(2k+1)\pi/2}$.
And that simplifies to $e^{ik\pi}=\{1,-1\}$.
This has connection to the anomally that
in general, $\log u+\log u\neq 2\log u$ for a complex numbers $u$.
