2008 AMC 12A Problem 23

Question

The solutions of the equation

z^4 + 4z^3 i - 6z^2 - 4zi - i = 0

are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

2^{\frac{5}{8}}

2^{\frac{3}{4}}

2

2^{\frac{5}{4}}

2^{\frac{3}{2}}

Add

1 + i

to both sides to recognize a perfect fourth power

The four values of

z + i

lie equally spaced on a circle of radius

2^{\frac{1}{8}},

forming a square

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.

Problem 23 in other years