According to the definition, $(z^\lambda)^\mu = \exp(\mu \log(z^\lambda))$ while $z^{\lambda \mu} = \exp(\lambda \mu \log(z))$. Now $\exp(a) = \exp(b)$ if and only if $a - b$ is an integer multiple of $2 \pi i$, so you want $\mu \log(z^\lambda) - \lambda \mu \log(z) = \mu (\log(z^\lambda) - \lambda \log(z))$ not to be an integer multiple of $2 \pi i$. Now $\log(z^\lambda) = \log(\exp(\lambda \log(z))) = \lambda \log(z) + 2 \pi i k$ where $k$ is the integer that puts the imaginary part of this in the interval $(-\pi, \pi]$, and then $\log(z^\lambda - \lambda \log(z) = 2 \pi i k$. So what you need is that $k$ is a nonzero integer and $\mu k$ is not an integer.
You could use any $z$ that is not $0$ or $1$. If $w = \log(z)$, choose a complex number $\lambda$ such that $\text{Im}(\lambda w)$ is outside the interval $(-\pi, \pi]$, find the integer $k$ such that $\text{Im}(\lambda w) - 2 \pi k$ is in $(-\pi, \pi]$, and then any $\mu$ such that $\mu k$ is not an integer.
Zev's example has $w = \pi i$ and $k = 1$ so his $\mu = 1/2$ works, but so would any non-integer $\mu$. Another example is $z = e$ with $\lambda = 2 \pi i$ and any non-integer $\mu$.