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​​.
