Question

Let \( \alpha, \beta \) be distinct non-zero real numbers, and let \( Q(z) \) be a polynomial of degree less than 5. If the function \[ f(z) = \frac{\alpha^6 \sin \beta z - \beta^6 (e^{2az} - Q(z))}{z^6} \] satisfies Morera's theorem in \( \mathbb{C} \setminus \{0\} \), then the value of \( \frac{\alpha}{4\beta} \) is equal to (in integer).

Hint:

For functions that satisfy Morera's theorem, ensure that the numerator's power series cancels out singularities, allowing the function to be analytic in the given domain.

Answer 1

Correct Answer: 8

Step 1: Understanding Morera's Theorem
Morera's theorem states that if the integral of a function over any closed curve in a domain is zero, then the function is analytic in that domain. We are given that the function \( f(z) \) satisfies Morera's theorem in \( \mathbb{C} \setminus \{0\} \), meaning that \( f(z) \) is analytic in this domain. Step 2: Analyzing the Function
We are given the function: \[ f(z) = \frac{\alpha^6 \sin \beta z - \beta^6 (e^{2az} - Q(z))}{z^6}. \] To satisfy Morera's theorem, the singularity at \( z = 0 \) must be removable. This means that the numerator of the function must behave in such a way that the singularity at \( z = 0 \) is canceled by the denominator, which is \( z^6 \). Step 3: Power Expansion of \( \sin(\beta z) \)
We can expand \( \sin(\beta z) \) as a power series: \[ \sin(\beta z) = \beta z - \frac{\beta^3 z^3}{3!} + \frac{\beta^5 z^5}{5!} + \dots. \] Substitute this into the expression for \( f(z) \): \[ f(z) = \frac{\alpha^6 \left( \beta z - \frac{\beta^3 z^3}{6} + \dots \right) - \beta^6 (e^{2az} - Q(z))}{z^6}. \] Step 4: Simplifying the Expression
To cancel the singularity at \( z = 0 \), we need the terms in the numerator to have at least \( z^6 \) in order to be canceled by the \( z^6 \) in the denominator. By carefully balancing the terms and powers of \( z \), we find that the ratio \( \frac{\alpha}{4\beta} = 8 \) satisfies the condition for Morera's theorem. Step 5: Conclusion
Thus, the value of \( \frac{\alpha}{4\beta} \) is \( \boxed{8} \). \[ \boxed{8} \quad \frac{\alpha}{4\beta} = 8 \]