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(B) Let $z = x + iy$. Since $B: \operatorname{Im}(z) = 0$, we have $y = 0$, so $|z| = |x|$.Substituting into the inequality for $A$: $\frac{1}{\log_2

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$(-\infty, -2) \cup (-1, 0) \cup (0, 1) \cup (2, \infty)$

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(B) Let $z = x + iy$. Since $B: \operatorname{Im}(z) = 0$, we have $y = 0$, so $|z| = |x|$.Substituting into the inequality for $A$: $\frac{1}{\log_2 |x|} - \frac{1}{\log_2 |x| - 1} Let $t = \log_2 |x|$. Then $\frac{1}{t} - \frac{1}{t-1} $\frac{(t-1) - t}{t(t-1)} $\frac{1}{t(t-1)} + 1 > 0 \Rightarrow \frac{1 + t^2 - t}{t(t-1)} > 0$.Since the discriminant of $t^2 - t + 1$ is $1 - 4 = -3 Thus, we require $t(t-1) > 0$, which implies $t  1$.Case $1$: $\log_2 |x| Case $2$: $\log_2 |x| > 1 \Rightarrow |x| > 2 \Rightarrow x \in (-\infty, -2) \cup (2, \infty)$.Combining these, $x \in (-\infty, -2) \cup (-1, 0) \cup (0, 1) \cup (2, \infty)$.

Consider the regions for complex number $z$ defined by $A: \frac{1}{\log_2 |z|} - \frac{1}{\log_2 |z| - 1} - 1 < 0$ and $B: \operatorname{Im}(z) = 0$. The range of values of $\operatorname{Re}(z)$ lying in the region $A \cap B$ is

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