Why (2i)^a/(-2i)^b is not equal to (2i)^a/(2i)^b * 1/(-1)^b?

Question

I'm trying to understand the following: why $\frac{(2i)^a}{(-2i)^b}$ cannot be represented as $\frac{(2i)^a}{(2i)^b}\frac{1}{(-1)^b}$ ? It works the other way: $\frac{(2i)^a}{(-2i)^b}=\frac{(-2i)^a}{(-2i)^b}(-1)^a$ - am I missing something or making a silly mistake here?

[edit] for integer a and b it does of course hold, but what if these are not integers? Initially I encountered the problem with a=43.5 and b=41.5, in which case my original equations yields 2, the second ("the other way") transformation yields also 2, but the first one (the problematic one) yields -2 (double checked in R and Mathematica).

If $a$ and $b$ are integers, I see no reason why it shouldn't work. Do you have a specific example? — Arthur, Nov 15 '16 at 14:05
I have edited my question with more details, briefly - it doesn't seem to hold for non-integer a,b. — szymekD, Nov 15 '16 at 17:06
Possible duplicate of Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? — Alex M., Nov 15 '16 at 17:15

score 1 · Answer 1 · answered Nov 15 '16 at 17:14

You are essentially asking whether $(-2 \Bbb i) ^a = (-1)^a (2 \Bbb i)^a$. For clarity, take $a = \frac 1 2$. The first question is: what do we mean by $x^{\frac 1 2}$ when $x$ is complex? Each complex number has two square roots, so in order for us to speak unambiguously about the complex square root we have to choose one branch of $\sqrt {\cdot}$, and once this branch has been chosen one has to cut the plane, and this introduces a number of problems (in particular, $\sqrt {uv}$ is not always equal to $\sqrt u \sqrt v$).

Why (2i)^a/(-2i)^b is not equal to (2i)^a/(2i)^b * 1/(-1)^b?

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