Determining holomorphicity I need to determine where the following function is differentiable and holomorphic in $\mathbb C$:
$$f(z)=(z-3)^i$$
I have the derivative as $df/dz= i(z-3)^{-1+i}$. 
The answer in my book says f is differentiable and holomorphic on $\mathbb C$ where $y\neq0$ and $x>3$. I don't see where this comes from. wolframalpha plotted the derivative and I can see that the imaginary part has a vertical asymptote at 0, but I can't see why f is not differentiable or holomorphic for $x\leq3$. 
How can I find the answer by looking at the function and its derivative?

The standard definition of $z^b$ for complex $b$ is $\exp(b \log z)$.  Different branches of the logarithm will give you different branches of $z^b$.  If you use the principal branch of the logarithm (which is what Wolfram Alpha uses, and apparently what your book uses), $\log z$ (and thus $z^b$) will be holomorphic in ${\mathbb C} \backslash (-\infty,0]$, and so $\log(z-3)$ and $(z-3)^b$ will be holomorphic in ${\mathbb C} \backslash (-\infty,3]$.  As $z$ crosses the line $(-\infty, 3)$, $\log(z-3)$ jumps by $2 \pi i$ so $z^b$ changes by a factor of $\exp(2 \pi i b)$.
