There is not an universally agreed-upon way to do this (raise an arbitrary complex number to the power 2.56). The usual way is to write the complex number $z$ in the form $r\exp(i\theta) = r(\cos\theta + i\sin\theta)$ where $r = |z| \geq 0$, $\theta$ is real (the equation is true due to Euler's Formula), then define
$$z^{2.256} = r^{2.56}\exp(2.56i\theta) = r^{2.56}(\cos(2.56\theta) + i\sin(2.56\theta))$$
(using Euler's formula for the second equation). The problem with this is that for any $z$, many choices of $\theta$ satisfy $z=r\exp(i\theta)$. If $z=0$, $\theta$ can be anything (but $0^{2.56}=0$, so there is no problem). If $z \neq 0$, the most popular choice is to use $\theta = \operatorname{Arg}(z)$, where $\operatorname{Arg}(z)$ is the unique angle in $(-\pi,\pi]$ satisfying $\cos\theta= Re(z)/|z|$ and $\sin\theta= Im(z)/|z|$. Note that it is a bad idea to say $\tan(\theta) = Im(z)/Re(z)$ because $Re(z)$ might be zero. It is even worse to define $\theta = \tan^{-1}(Im(z)/Re(z))$ because not only can $Re(z)$ be zero, but this way, $\theta$ must be in a right-hand quadrant, which might not be correct.
While using $\operatorname{Arg}(z)$ is the most popular way to choose $\theta$, I don't know if there is any mathematical reason why it is the best way.
To compute $\operatorname{Arg}(z)$ using Java, if you are lucky, there will be a two-argument "Arg"-type function. If this does not exist, you should use the regular inverse tangent function, but you will have to do several cases, depending on the signs of $x$ and $y$, and in some cases you will have to add or subtract $\pi$ from the output of the inverse tangent function.
EDIT: dfeuer makes a good observation in his/her answer, and I hope s/he doesn't mind if I steal it and repeat it here: if $n$ is an integer and $z$ a complex number, then $z^n$ is unambiguously defined and can be computed in the obvious way (repeated multiplication and/or repeated squaring) or by using the scheme above (you will get the same answer regardless of which sensible $\theta$ you use). Of course, $0^n$ is undefined if $n<0$, and it is debatable whether $0^0$ should be given a value (in many applications, people seem to treat it as $1$). If the exponent $\alpha$ is rational, then as s/he writes, then $z^\alpha$ has finitely many values that might make sense. If your exponent is 2.56, which simplfies to $64/25$, there are "only" 25 possible possible values for $z^{2.56}$ if $z \neq 0$.
Best Answer
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = \sqrt2 e^{-i\pi/4}.$ Hence $$(1 - i)^9 = 2^{9/2} e^{-i9\pi/4} = 16\sqrt2 e^{-i\pi/4} = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that computing $\cos(3\arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.