Range of Modulus Expression for Complex Numbers in Given Region
Let be any point in and let be any point satisfying
Then, lies between
This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.
Options
and
and
and
and
Topics & Concepts
Step-by-Step Solution
To find the range of the expression , we recall the definitions of the sets from the passage:
Step 1: Finding the bounds on
The set represents a closed disc centered at with radius .
The distance of the center from the origin is:
For any point , must belong to , which means:
Since , we have , which implies:
It can also be checked that points on the boundary with satisfy the conditions of and (for instance, , where and ). Thus:
Step 2: Finding the bounds on
The point satisfies:
This is the open disc of radius centered at . Since the distance of the center from the origin is , the origin lies inside this disc. Therefore:
Multiplying by :
Step 3: Determining the range of
Adding the inequalities:
Approximating :
Thus, the exact range of values that the expression can attain is a sub-interval of:
Step 4: Analysis of Options
Depending on the interpretation of "lies between" (whether finding a containment interval or estimating extreme bounds):
- The interval is contained within (Option D).
- Using loose triangle inequalities like or , Options B and C are also frequently interpreted as valid bounds.
Due to multiple valid interpretations of the loose bounds, options B, C, and D are accepted.