To find the set of points formed by the expression π<|π§β4+2π|<3π, we need to solve the inequality.
Let π§ = π₯ + π¦π, where π₯ and π¦ are real numbers.
The inequality can be written as:
π < |(π₯ + π¦π) β (4 + 2π)| < 3π
Simplifying the expression inside the absolute value:
π < |π₯ β 4 + (π¦ β 2)π| < 3π
Let π = π₯ β 4 and π = π¦ β 2. The inequality becomes:
π < |π + ππ| < 3π
Using the polar representation of complex numbers, π + ππ can be written as:
π + ππ = π(πππ (π) + ππ ππ(π))
where π = |π + ππ| is the magnitude of π + ππ and π is the argument of π + ππ.
We can rewrite the inequality as:
π < |π(πππ (π) + ππ ππ(π))| < 3π
Since π is always nonnegative, we can remove the absolute value and rewrite the inequality as:
π < π < 3π
Therefore, the set of points formed by the expression π < |π§β4+2π| < 3π is the set of all complex numbers π§ such that the magnitude of π§ β (4 + 2π) is between π and 3π.
Answer: The set of points formed by the expression π < |π§β4+2π| < 3π is π < π < 3π, where π§ = π₯ + π¦π and π is the magnitude of π§ β (4 + 2π).