Problem:
There are 2n complex numbers that satisfy both z28−z8−1=0 and ∣z∣=1. These numbers have the form zm=cosθm+isinθm. where 0≤θ1<θ2<…<θ2n<360 and angles are measured in degrees. Find the value of θ2+θ4+⋯+θ2n.
Solution:
Because ∣z∣=1,z=cosθ+isinθ, with 0≤θ<360. Now cos28θ−cos8θ=1 and sin28θ−sin8θ=0. From the latter equation, conclude that 28θ+8θ=180+360k for some integer k, so θ=10k+5. It follows from the former equation that
−2sin228θ+8θsin228θ−8θ=1
which is equivalent to sin18θsin10θ=−21. Substitute θ=10k+5 to obtain sin(180k+ 90)sin(100k+50)=−21, or sin(100k+50)=(−1)k+1⋅21. When k=2m−1 (that is, when k is odd )sin(200m−50)=21. Then 200m−50≡30 or 150(mod360) yields m≡4 or 1(mod9),k≡7 or 1(mod18), and θ≡75 or 15(mod180). When k=2m. similar reasoning leads to θ≡165 or 105(mod180). Thus θ≡±15(mod90) and θ2+θ4+θ6+θ8=Σk=14(90k−15)=840.
OR
Let cis θ denote cosθ+isinθ. If z satisfies z28−z8−1=0, then z8(z20−1)=1. Because ∣z∣=1, it follows that ∣∣∣z20−1∣∣∣=1 and ∣∣∣z20∣∣∣=1. This can happen only if z20=cis(±60), in which case z20−1=cis(±120). Therefore z8=cis(∓120) and z4=z20/(z8)2=cis(±300)=cis(∓60), which implies that z=cis(90k∓15) for some integer k. Such z satisfy z28=cis(∓60)=1+cis(∓120)=1+z8. Thus the equation z28−z8−1=0 has eight solutions on the unit circle, namely θ1=15,θ2=75,θ3=105. θ4=165,θ5=195,θ6=255,θ7=285, and θ8=345. It follows that θ2+θ4+θ6+θ8=840.
OR
With z=cosθ+isinθ, write the equation as z18(z10−z−10)=1, and notice that z10−z−10=2isin10θ. Then, taking the absolute value of each side of the original equation, 2sin10θ=±1, and thus z is a solution with ∣z∣=1 if and only if
z18=∓i and sin10θ=±1/2
Hence the desired values of θ satisfy 18θ≡270 or 90(mod360) with 10θ≡30 or 150 (mod360) in the first case and 10θ≡210 or 330(mod360) in the second. Thus
θ≡15(mod20) and θ≡3 or 15(mod36)
or
θ≡5(mod20) and θ≡21 or 33(mod36)
The smallest positive solutions are 75 and 15 in the first case, and 165 and 105 in the second. Solutions are congruent modulo 180, so the solutions between 0 and 360 are 15, 75,105,165,195,255,285,345. Thus θ2+θ4+θ6+θ8=840.
The problems on this page are the property of the MAA's American Mathematics Competitions