Question
If , then the maximum value of is :
Options
Solution
Key Concepts and Formulas
- Geometric Interpretation: represents the distance between complex numbers and in the Argand plane. represents all points within or on a circle of radius centered at .
- Triangle Inequality: For complex numbers and , . Equality holds if and only if and have the same argument or either is zero.
Step-by-Step Solution
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Step 1: Interpret the Given Condition Geometrically The condition can be rewritten as . This represents all complex numbers lying within or on a circle centered at (i.e., the point in the complex plane) with a radius of .
- Why? Rewriting in the form allows us to directly identify the center and radius, which are crucial for the geometric approach.
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Step 2: Interpret the Expression to Maximize Geometrically We want to find the maximum value of , which can be rewritten as . This represents the distance between the complex number and the point (i.e., the point in the complex plane).
- Why? This allows us to frame the problem as finding the point within the circle defined in Step 1 that is farthest from the point .
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Step 3: Geometric Solution - Find the Point z for Maximum Distance The maximum distance between the point and any point within the circle will occur when lies on the circle's boundary and is collinear with the center of the circle and the point , on the opposite side of the center from .
- Why? This is a fundamental geometric principle: to maximize the distance from a point to a circle, we must extend a line from the point through the center to the far side of the circle.
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Step 4: Geometric Solution - Calculate the Maximum Distance The distance between the center of the circle, , and the point is . The maximum distance from to any point within the circle is the sum of this distance and the radius of the circle: .
- Why? We are simply adding the distance from the fixed point to the circle's center to the radius to find the maximum distance to a point on the circle.
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Step 5: Algebraic Solution - Apply the Triangle Inequality We want to maximize . Rewrite as . Then, by the triangle inequality, .
- Why? Rewriting in terms of allows us to use the given condition . The triangle inequality gives us an upper bound for .
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Step 6: Algebraic Solution - Substitute the Given Condition and Simplify Since , we have . Therefore, .
- Why? Substituting the given condition into the inequality allows us to find the maximum possible value of .
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Step 7: Algebraic Solution - Verify the Equality Condition The maximum value of 6 is achieved when and the arguments of and are equal. This occurs when , which gives . When , .
- Why? We need to ensure that the upper bound we found using the triangle inequality is actually attainable. We do this by finding a value of that satisfies both the given condition and achieves the maximum value.
Common Mistakes & Tips
- Center Identification: Be careful with the sign when identifying the center from . In , the center is at , not .
- Equality Condition: Remember that the triangle inequality becomes an equality when the complex numbers involved have the same argument (direction).
Summary
The problem asks for the maximum value of given . Both geometric and algebraic approaches show that the maximum value is 6. The geometric approach visualizes the problem as finding the farthest point within a circle from a given point. The algebraic approach uses the triangle inequality to find an upper bound for and then verifies that this upper bound is attainable.
The final answer is . which corresponds to option (A).