Complex Conjugates and Inequality
In the complex plane, if |z| > 3, where z is a complex number, show that its conjugate, θ, must satisfy |zθ| > 9. Assume z = a + bi, where a and b are real numbers.
1 Answer
📌 CONCEPT: The complex conjugate of a complex number z = a + bi is given by θ = a - bi.
📐 RULE / FORMULA: We know that the modulus of the product of two complex numbers is the product of their moduli, i.e., |zθ| = |z||θ|.
💡 WORKED EXAMPLE: Let's consider a complex number z = 4 + 3i. Its conjugate is θ = 4 - 3i. Given |z| > 3, i.e., |4 + 3i| > 3, we need to show that |zθ| > 9. Using the rule, |zθ| = |z||θ| = |4 + 3i||4 - 3i| = √(4^2 + 3^2) * √(4^2 + (-3)^2) > 3√(16 + 9) = 3√25 = 15 > 9. Hence, |zθ| > 9.
⚠️ COMMON MISTAKE: Students often forget to consider the absolute value when multiplying complex numbers, leading to incorrect results.
17 Oct 26
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