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?

$2^{5/8}$

$2^{3/4}$

$2$

$2^{5/4}$

$2^{3/2}$

Answer: D

Concepts:binomial theorem complex number roots of unity

Difficulty rating: 2240

Solution:

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 $|1 + i|^{1/4} = (2^{1/2})^{1/4} = 2^{1/8},$ and they form a square. Subtracting $i$ merely translates it.

A square inscribed in a circle of radius $2^{1/8}$ has diagonal $2 \cdot 2^{1/8} = 2^{9/8},$ so its side is $\tfrac{2^{9/8}}{\sqrt{2}} = 2^{5/8}.$

The area is $\left(2^{5/8}\right)^2 = 2^{5/4}.$

Thus, D is the correct answer.

2008 AMC 12A Problem 23

Answer: D

Difficulty rating: 2240

Solution:

Problem 23 in Other Years