Problem:
Let P(z)=z8+(43+6)z4−(43+7). What is the minimum perimeter among all the 8-sided polygons in the complex plane whose vertices are precisely the zeros of P(z) ?
Answer Choices:
A. 43+4
B. 82
C. 32+36
D. 42+43
E. 43+6 Solution:
Factoring or using the quadratic formula with z4 as the variable yields P(z)=(z4−1)(z4+(43+7)). Moreover, 43+7=(3+2)2 and 2(3+2)=23+4=(3+1)2; thus 43+7=(21(6+2))4. If w=21(3+ 1 ), then the eight zeros of P(z) are 1,−1,i,−i,w(1+i),w(−1+i),w(−1−i), and w(1−i).
The distances from 1 to the other zeros are
∣1−(−1)∣=2,∣1±i∣=2,∣1−w(1±i)∣=(1−w)2+w2=2, and ∣1−w(−1±i)∣=(1+w)2+w2=23+4=3+1
Similarly, the distances from w(1+i) to the other zeros are
∣w(1+i)−1∣=∣w(1+i)−i∣=2, and ∣w(1+i)+1∣=∣w(1+i)+i∣=3+1
Because the set of zeros is 4 -fold symmetric with respect to the origin, it follows that every line segment joining two of the zeros has length at least 2. This shows that any polygon with vertices at the zeros has perimeter at least 82. Finally, note that the polygon with consecutive vertices 1,w(1+i),i,w(−1+i), −1,w(−1−i),−i, and w(1−i) has perimeter 82.
