# Complex exponentiation

1. Jun 25, 2011

### 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?

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

2. Jun 25, 2011

### 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.

3. Jun 25, 2011

### bruno67

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

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

4. Jun 25, 2011

Indeed!