# How does one find $z,w,\lambda \in \mathbb{C}$ such that $(zw)^{\lambda}\neq z^{\lambda}w^{\lambda}$?

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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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

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.