Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to find $z,w,\lambda \in \mathbb{C}$ such that $(zw)^{\lambda}\neq z^{\lambda}w^{\lambda}$. I wasn't able to find an example so far.

If you take $z$ and $z^{-1}$, then $(zw)^{\lambda}=z^{\lambda}w^{\lambda}$. If $z=1+i, w=-1+1$ and $\lambda =i$ it works out too.

The authors are defining $z^{\lambda}$ as $e^{\lambda \operatorname{Log}(z)}$.

I would appreciate any hint!

share|improve this question
1  
Even for something as simple as $\lambda=1/2$, there is in general more than one object that could be called $z^{\lambda}$. By equality do you mean equality as sets? –  André Nicolas Mar 5 '12 at 23:50
    
How are you defining a complex power? –  Juan S Mar 5 '12 at 23:51
    
And does 'log' mean principal logarithm? –  Juan S Mar 6 '12 at 0:07
    
@AndréNicolas: The author defines $z^{\lambda}$ as $e^{\lambda \operatorname {Log}(z)}$. –  spohreis Mar 6 '12 at 0:09
    
@JuanS: Sorry, but I was wrong before. That's the reason I deleted my previuos comment. –  spohreis Mar 6 '12 at 0:10
add comment

1 Answer

up vote 4 down vote accepted

One simple counterexample is $z=w=-1$ and $\lambda=1/2$. Then we have $$(zw)^\lambda=(-1\cdot -1)^{1/2} = 1^{1/2} = 1$$ but $$z^\lambda w^\lambda = (-1)^{1/2}(-1)^{1/2} = i\cdot i = -1$$ because $(-1)^{1/2} = e^{\frac{\operatorname{Log}(-1)}2} = e^{\frac{\pi i}2} = i$.

Notice that, for $\mu=2$, this also shows that $(z^\mu)^\lambda \ne (z^\lambda)^\mu$ in general when $\mu$ and/or $\lambda$ is not an integer and the powers are defined through the principal logarithm (or any single-valued logarithm, really) -- and further that $(z^\lambda)^\mu\ne z^{\lambda\mu}$ sometimes.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.