Problem:
Let f:C→C 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={z∈C:Im(z)>0}. If z1,z2∈H and f(z1)=f(z2), then z12−z22+i(z1−z2)=(z1−z2)(z1+z2+i)=0. Because Im(z1)>0 and Im(z2)>0, it follows that z1+z2+i=0. Thus z1=z2; 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(−r)=r2+1−ir describes the bottom part of the parabola x=y2+1. Because f(i)=−1, it follows that the image set f(H) equals {w∈C:Re(w)<(Im(w))2+1}. Thus the set of complex numbers w∈f(H) with integer real and imaginary parts of absolute value at most 10 is equal to
S={w=a+ib∈C:a,b∈Z,∣a∣≤10,∣b∣≤10, and a<b2+1}
Because f is one-to-one, the required answer is ∣∣∣f−1(S)∣∣∣=∣S∣ and
12−b=−3∑3a=b2+1∑101=441−b=−3∑3(10−b2)41−(1+6+9+10+9+6+1)=399.
The problems on this page are the property of the MAA's American Mathematics Competitions