Problem:
The solutions of the equation z4+4z3iβ6z2β4ziβi=0 are the vertices of a convex polygon in the complex plane. What is the area of the polygon?
Answer Choices:
A. 285β
B. 243β
C. 2
D. 245β
E. 223β Solution:
Adding 1+i to each side of the given equation gives
wkβ=21/8(cos(16Οβ+2Οβk)+isin(16Οβ+2Οβk)) for k=0,1,2, or 3
and the four solutions for z=wβi are
zkβ=21/8(cos(16Οβ+2Οβk)+isin(16Οβ+2Οβk))βi for k=0,1,2, or 3
Note that w0β,w1β,w2β, and w3β are equally spaced around the circle of radius 21/8 centered at (0,0), so z0β,z1β,z2β, and z3β are equally spaced around the circle of radius 21/8 centered at (0,β1). Therefore z0β,z1β,z2β, and z3β are vertices of a square with side length 21/82β=25/8 and area (25/8)2=25/4β.
Let a satisfy a4=1+i, and let w=z+i. Then w4=a4, so the possible values for w are a,ia,βa, and βia, which are the vertices of a square with diagonal 2β£aβ£=282β. The transformation w=z+i is a translation, so it preserves area. Hence the area of the original polygon is (282β)2/2=242β=25/4β.