2008 AMC 12A Problem 23

Adding $1 + i$ to both sides, the left side becomes $$\begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4, \end{aligned}$$ so $(z + i)^4 = 1 + i.$ The four solutions for $w = z + i$ are equally spaced on a circle of radius $|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}},$ and they form a square. Subtracting $i$ merely translates it. A square inscribed in a circle of radius $2^{\frac{1}{8}}$ has diagonal $2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}},$ so its side is $\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}.$ The area is $$\left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}.$$ Thus, D is the correct answer.