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:
Using the fact zzΛ=β£zβ£2, the equation rewrites itself as
2(z+2)(zΛ+2)+(z2+1)(zΛ2+1)+31β12zzΛ+2zzΛ+4(z+zΛ)+8+z2zΛ2+(z2+zΛ2)+32((z2+2zzΛ+zΛ2)+4(z+zΛ)+4)+(z2zΛ2β12zzΛ+36)(z+zΛ+2)2+(zzΛβ6)2β=12zzΛ=0=0=0β
As the two quantities in the parentheses are real, both quantities must equal 0 so
z+z6β=z+zΛ=(A) β2β
OR
Let z=a+bi,z2=a2βb2+2abi. Using the equation given in the problem,
2((a+2)2+b2)+((a2βb2+1)2+(2ab)2)+312a2+8a+8+2b2+a4+b4+1+2a2β2b2β2a2b2+4a2b2+31a4+b4β8a2β12b2+2a2b2+8a+40(a2+b2)2β12(a2+b2)+4(a2+2a+1)+36(a2+b2β6)2+4(a+1)2β=12(a2+b2)=12a2+12b2=0=0=0β
Therefore, a2+b2β6=0 and a+1=0.
ab2bβ=β1=6β1=5=5ββ
z+z6ββ=a+bia2βb2+6+2abiβ=β1+i5β1β5+6+2(β1)5βiβ=β1+i5β2β2i5ββ=(A) β2ββ
The problems on this page are the property of the MAA's American Mathematics Competitions