What combinations satisfy this complex exponentiation property?

[bruno67]

Suppose we have the product

$$[(\pm ia) (\pm ib)]^{-\alpha}$$
where$a, b, \alpha >0$. For which of the combinations (+,+), (+,-), (-,+), and (-,-) is the following property satisfied?

$$[(\pm ia) (\pm ib)]^{-\alpha}=(\pm ia)^{-\alpha} (\pm ib)^{-\alpha}$$

 

[micromass]

Hi Bruno67!

Using the definition of exponentiation we have that

$$[(\pm ia)(pm ib)]^{-\alpha}=e^{-\alpha Log((\pm ia)(\pm ib))}$$

So the question becomes when

$$Log((\pm ia)(\pm ib))=Log(\pm ia)+Log(\pm ib)$$

Solve this using the definition of the logarithm.

 

[bruno67]

Thanks, so it holds in all cases except the (-,-) one. In that case we have

$$[(-ia) (-ib)]^\alpha = (-ia)^\alpha (-ib)^\alpha (-1)^{2\alpha}.$$

 

[micromass]

Indeed!

 

[CellCoree]

The property is satisfied for all combinations (+,+), (+,-), (-,+), and (-,-). This is because when we expand the expression on the right side of the equation, we get (\pm ia)^{-\alpha} (\pm ib)^{-\alpha} = (\pm 1)^{-\alpha} (a)^{-\alpha} (\pm 1)^{-\alpha} (b)^{-\alpha} = 1 (a)^{-\alpha} 1 (b)^{-\alpha} = (a)^{-\alpha} (b)^{-\alpha}. This is the same as the expression on the left side, so the property is satisfied for all combinations.