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 $10?$

$399$

$401$

$413$

$431$

$441$

Answer: A

Concepts:complex number lattice point

Difficulty rating: 2790

Solution:

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.

2013 AMC 12A Problem 25

Answer: A

Difficulty rating: 2790

Solution:

Problem 25 in Other Years