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?

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

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: ${\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cdot i \cdot \cos \frac {\pi}{4} + 2^{\frac{1}{2}}\cdot \sin \frac {\pi}{4}$ 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 $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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2008 AMC 12A Problems/Problem 23

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