Problem 19 Complex Numbers

Problem 19: Complex Numbers

AMC 12 2008A #23

Problem: 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? Solution: We can look at the symmetry of the coefficients 1, 4, 6, 4, 1. This immediately reminds us of the binomial expansion of (x + 1)4 . However, we have to adjust this for the i’s and such. Notice that the complex units of the coefficients are 1, i, −1, −i, −i. We notice that this is close to the pattern formed by 1, i, i2 , i3 , i4 . Based on these two observations of symmetry, we can deduce that (z + i)4 is very close to the given equation here. More specifically, we have that (z + i)4 = 1 + i =

√

2 · cis

π 4

A translation of z by −i simply means replacing z with z − i. Remember that the imaginary part is in the y-coordinate direction. So if we reduce z by the value i, visually, it’s the same as sliding z downwards by 1 unit in the y direction. If you slide a shape down one unit, its area doesn’t change. So now, we started from (z + i)4 =

√

2 · cis

π 4




Now, we’ll replace z with z − i, and we get (z − i + i)4 = z 4 =

√

2 · cis

π 4




.

We can also rotate by an angle of -45 degrees counterclockwise, or −π/4 in radian measure, and the area still won’t change. To rotate a complex number around the origin by an angle of θ, you multiply it by cos θ + i sin θ, abbreviated cis θ. So let’s do that: z 4 =

√ 2 · cis π4 · cis −π 4

So now, the two rotations cancel out, since they are in opposite directions, and you get that the original area is the same as the area of the quadrilateral whose vertices are roots of √ z4 = 2 Now, if we solve that equation, we get the numbers

√ 8

√ √ √ 2, − 8 2, i 8 2, −i 8 2.

√ 4 If we plot those numbers as points in the complex plane, we find a square with area 2 2 .

Solution was combined from two solutions initially written by Alex Anderson & The Zuton Force and compiled from Art of Problem Solving Forums.