Problem:
Let z be a complex number satisfying 12β£zβ£2=2β£z+2β£2+β£β£β£βz2+1β£β£β£β2+31. What is the value of z+z6β?
Answer Choices:
A. β2
B. β1
C. 21β
D. 1
E. 4 Solution:
Answer (A): Recall that if w is a complex number and wΛ denotes its conjugate, then wβ wΛ=β£wβ£2. Using this relationship and properties of conjugation, the given equation can be rewritten as
12zzΛ=2(z+2)(zΛ+2)+(z2+1)(zΛ2+1)+31
This is equivalent to 12zzΛ=2zzΛ+4z+4zΛ+8+(zzΛ)2+z2+zΛ2+1+31. Rearranging the terms gives
((zzΛ)2β12zzΛ+36)+(z2+zΛ2+2zzΛ+4z+4zΛ+4)=0
This implies that (zzΛβ6)2+(z+zΛ+2)2=0. Because zzΛ and z+zΛ are both real numbers, it must be that zzΛβ6=0 and z+zΛ+2=0. Thus zzΛ=6 and z+zΛ=β2, so z+z6β=z+zΛ=β2.
Note: The values of z that satisfy the given equation are β1Β±i5β, for which each side equals 72 .