2013 AMC 12A Problem 25

Question

Let

f : \mathbb{C} \to \mathbb{C}

be defined by

f(z) = z^2 + iz + 1.

How many complex numbers

z

are there such that

\operatorname{Im}(z) \gt 0

and both the real and the imaginary parts of

f(z)

are integers with absolute value at most

10?

399

401

413

431

441

Show

f

is one-to-one on the upper half-plane, so counting

z

equals counting valid

f(z)

The image is

\{w : \operatorname{Re}(w) \lt (\operatorname{Im}(w))^2 + 1\};

count lattice points with

|\operatorname{Re}|, |\operatorname{Im}| \le 10

On the upper half-plane

H,

if

f(z_1) = f(z_2)

then

(z_1 - z_2)(z_1 + z_2 + i) = 0;

since

\operatorname{Im}(z_1), \operatorname{Im}(z_2) \gt 0,

the factor

z_1 + z_2 + i \ne 0,

so

f

is one-to-one on

H.

For real

r,

the boundary values

f(r)=r^2+1+ir

trace the parabola

\operatorname{Re}(w)=(\operatorname{Im}(w))^2+1.

Since

f(i)=-1

lies to its left and

f

is continuous and one-to-one on

H,

its image consists precisely of the values

w

satisfying

\operatorname{Re}(w)<(\operatorname{Im}(w))^2+1.

Thus we count

w = a + ib

with

a, b \in \mathbb{Z},

|a|, |b| \le 10,

and

a \lt b^2 + 1:

\begin{gathered} |S| = 21^2 \\ {}- \sum_{b=-3}^{3}(10 - b^2) \\ = 441 - 42 = 399. \end{gathered}

Thus, the correct answer is A.

Problem 25 in other years