Question
If is a complex number such that , then the minimum value of is :
Options
Solution
Key Concepts and Formulas
- Geometric Interpretation: represents the distance between complex numbers and in the complex plane. represents the distance of from the origin.
- Triangle Inequality: For complex numbers and , and .
- Modulus of a Complex Number: If , then .
Step-by-Step Solution
Step 1: Rewrite the Expression and Identify the Fixed Point
We are given the expression and want to find its minimum value, subject to . We can rewrite the expression as , where . Therefore, .
Reasoning: Rewriting the expression in the form highlights that we are finding the distance between the complex number and a fixed point . This is essential for a geometric approach.
Step 2: Calculate the Modulus of the Fixed Point
We calculate to determine its distance from the origin: Thus, .
Reasoning: Calculating tells us the location of the fixed point relative to the region where can lie.
Step 3: Analyze the Region for z
We are given that . This means that lies within or on the unit circle centered at the origin. Since , the point lies outside the unit circle.
Reasoning: Since the fixed point is outside the region where can be, the minimum distance between and will not be 0. It will occur when is on the boundary of the region, i.e., when .
Step 4: Apply the Triangle Inequality
We want to minimize . Using the triangle inequality , we let and : Substitute :
Reasoning: The triangle inequality provides a lower bound for the distance . To minimize , we minimize this lower bound, subject to the constraint .
Step 5: Minimize the Lower Bound
We want to minimize subject to . Let . Then we want to minimize where . Since , the closest value to in the interval is . Therefore, the minimum value of occurs when . Substituting , we get: Minimum value .
Reasoning: By considering the range of possible values for and the properties of the absolute value function, we find the value of that minimizes the lower bound.
Step 6: Geometric Interpretation (Verification)
The minimum distance from any point within or on the unit circle to the fixed point occurs when is on the line connecting the origin to and . The minimum distance is .
Reasoning: This geometric argument confirms our algebraic result, providing a clear visual understanding.
Conclusion
The minimum value of is .
Common Mistakes & Tips:
- Sign Errors: Be careful with signs when identifying the fixed point . The expression represents the distance between and .
- Triangle Inequality Form: Choose the appropriate form of the triangle inequality based on whether you are finding a maximum or a minimum value. For minimums, is usually most effective.
- Geometric Visualization: Always try to visualize complex number problems geometrically. The modulus is a distance, and the condition means is inside or on a circle of radius centered at the origin.
Summary
This problem demonstrates how to combine geometric intuition with the triangle inequality to find the minimum value of an expression involving complex numbers. By identifying the fixed point, its location relative to the allowed region for , and using the appropriate form of the triangle inequality, we systematically determined that the minimum value is .
The final answer is \boxed{\frac{3}{2}}, which corresponds to option (C).