2008 AMC 12A Problems/Problem 23

Problem

The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

$\textbf{(A)}\ 2^{\frac{5}{8}} \qquad \textbf{(B)}\ 2^{\frac{3}{4}} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2^{\frac{5}{4}} \qquad \textbf{(E)}\ 2^{\frac{3}{2}}$

Solution

Looking at the coefficients, we are immediately reminded of the binomial expansion of ${\left(x+1\right)}^{4}$ .

Modifying this slightly, we can write the given equation as: ${\left(x+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cdot \text{cis}\, \frac {\pi}{4}$ We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: $z^{4}=2^{\frac{1}{2}}$

Because the roots of this equation are created by rotating $\frac{\pi}{2}$ radians successively about the origin, the quadrilateral is a square.

We know that half the diagonal length of the square is ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$

Therefore, the area of the square is: $\frac{\left( 2 \cdot 2^{\frac{1}{8} \right)}^{2}}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$ (Error compiling LaTeX. Unknown error_msg)

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2008 AMC 12A Problems/Problem 23

Problem

Solution

See Also