Problem:
Let f:CβC be defined by f(z)=z2+iz+1. How many complex numbers z are there such that Im(z)>0 and both the real and the imaginary parts of f(z) are integers with absolute value at most 10 ?
Answer Choices:
A. 399
B. 401
C. 413
D. 431
E. 441
Solution:
Let H={zβC:Im(z)>0}. If z1β,z2ββH and f(z1β)=f(z2β), then z12ββz22β+i(z1ββz2β)=(z1ββz2β)(z1β+z2β+i)=0. Because Im(z1β)>0 and Im(z2β)>0, it follows that z1β+z2β+iξ =0. Thus z1β=z2β; that is, the function f is one-to-one on H. Let r be a positive real number. Note that f(r)=r2+1+ir describes the top part of the parabola x=y2+1. Similarly, f(βr)=r2+1βir describes the bottom part of the parabola x=y2+1. Because f(i)=β1, it follows that the image set f(H) equals {wβC:Re(w)<(Im(w))2+1}. Thus the set of complex numbers wβf(H) with integer real and imaginary parts of absolute value at most 10 is equal to
S={w=a+ibβC:a,bβZ,β£aβ£β€10,β£bβ£β€10, and a<b2+1}
Because f is one-to-one, the required answer is β£β£β£βfβ1(S)β£β£β£β=β£Sβ£ and
12βb=β3β3βa=b2+1β10β1=441βb=β3β3β(10βb2)41β(1+6+9+10+9+6+1)=399β.β
The problems on this page are the property of the MAA's American Mathematics Competitions