I would like to show that $\displaystyle \left|\int_C{{z^{-4}}} ~dz \right| \le 4\sqrt2$ where $C$ is ANY path between $z = i$ and $z = 1$.
Show that a path $C$ where the length of that path is infinity or approaches infinity produces a maximum approaching $4 \sqrt2$. I haven't the slightest clue how to construct an infinitely long path.
Since the pole of $z^{-4}$ has residue $0$, the integral is in fact path-independent.
Also, you can write down an explicit antiderivative. You do have to assume the path does not hit $0$.
EDIT: The second part is simply false. The integral on any path from $i$ to $1$ is $(i-1)/3$ which has absolute value $\sqrt{2}/3$. There is no way to get anything close to $4 \sqrt{2}$.