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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!

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    @JuanS: Sorry, but I was wrong before. That's the reason I deleted my previuos comment.2012-03-06

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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.