Task B-2.1. How many complex numbers $z=a+bi$ are there for which:

$$
a, b \in \mathbb{Z}, \quad a \cdot b \geq 0 \quad \text { and } \quad \frac{|z|-16}{1-|z|} \geq 2 ?
$$
## Solution.

Let's write the given inequality in a more convenient form.

$$
\begin{aligned}
& \frac{|z|-16}{1-|z|} \geq 2 \\
& \frac{|z|-16}{1-|z|}-2 \geq 0 \\
& \frac{3(|z|-6)}{1-|z|} \geq 0
\end{aligned}
$$

The solution to this inequality is equivalent to the solution of the quadratic inequality

$$
(|z|-6)(1-|z|) \geq 0
$$

with the condition $|z| \neq 1$, and its solution is the interval $\langle 1,6]$. Therefore, $1<|z| \leq 6$, meaning the solution to the inequality consists of all points within the annulus with radii 1 and 6. We need to count how many points with integer coordinates in this annulus belong to the first and third quadrants.

![](https://cdn.mathpix.com/cropped/2024_05_30_6bd7d8198e5de61c26bbg-06.jpg?height=710&width=786&top_left_y=1946&top_left_x=645)

It is sufficient to count the points $(a, b)$ for which $a, b \in \mathbb{N} \cup\{0\}$ and $1<a^{2}+b^{2} \leq 36$.

If $a=0$, then $b^{2} \leq 36$, i.e., $b \in\{2,3,4,5,6\}$, which gives a total of 5 points.

If $a=1$, then $b^{2} \leq 35$, i.e., $b \in\{1,2,3,4,5\}$, which gives a total of 5 points.

If $a=2$, then $b^{2} \leq 32$, i.e., $b \in\{0,1,2,3,4,5\}$, which gives a total of 6 points.

If $a=3$, then $b^{2} \leq 27$, i.e., $b \in\{0,1,2,3,4,5\}$, which gives a total of 6 points.

If $a=4$, then $b^{2} \leq 20$, i.e., $b \in\{0,1,2,3,4\}$, which gives a total of 5 points.

If $a=5$, then $b^{2} \leq 11$, i.e., $b \in\{0,1,2,3\}$, which gives a total of 4 points.

If $a=6$, then $b^{2} \leq 0$, i.e., $b=0$, which gives 1 point.

In total, we have $2 \cdot(5+5+6+6+5+4+1)=64$ points, or 64 complex numbers with the given properties.