Question
If , satisfies the equation , then is equal to :
Options
Solution
Key Concepts and Formulas
- Modulus of a complex number: If , then .
- Properties of the modulus:
- Given , then and , which implies and thus .
Step-by-Step Solution
1. Rearrange the given equation. We are given . To prepare for applying the modulus, isolate the terms: Explanation: Isolating the terms involving on one side allows us to apply the modulus operation to both sides and simplify the equation.
2. Apply the modulus to both sides of the equation. Take the modulus of both sides: Explanation: Applying the modulus to both sides allows us to work with magnitudes, transforming the complex equation into a real-valued equation.
3. Simplify the moduli using relevant properties.
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Left Hand Side (LHS): Using the property : Explanation: This simplifies the LHS to the square of the modulus of .
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Right Hand Side (RHS): Using the property : Using : Using : Therefore, the RHS simplifies to . Explanation: We systematically applied the modulus properties. The modulus of a product was broken down, the modulus of was found to be 1, and the modulus of the conjugate was replaced with the modulus of , simplifying the entire RHS to .
4. Equate the simplified expressions and solve for . From Step 3, we have . Rearrange the equation: Factor out : This gives us two possible solutions for : Explanation: We transformed the equation into a solvable form involving , leading to two potential values.
5. Validate the solutions for using the given conditions. The problem states that with . If , then and . If , then , which implies . This can only be true if and . However, and contradicts the given condition . Therefore, is not a valid solution. The only valid solution is . Explanation: The condition means cannot be the origin (). Since represents the distance from the origin, would mean , which is disallowed.
6. Calculate the required value . The question asks for the value of . Using the property , and since we found : Thus, .
Common Mistakes & Tips
- Always check given conditions: The condition was crucial here to eliminate the extraneous solution .
- Modulus properties: Memorize and understand how to apply the modulus properties to simplify calculations.
- Distinguish between and : is a complex number, while is its magnitude (a non-negative real number).
Summary
This problem demonstrates how to solve complex number equations by applying the modulus operation and leveraging the properties of complex moduli. We transformed the complex equation into a real-valued equation, solved for , and then used the given condition to discard the extraneous solution. The final answer is .
Final Answer
The final answer is \boxed{1}, which corresponds to option (D).