Question

The values of \( a \) and \( b \) for the function \( f(z) = (x^2 + a y^2 - 2xy) + i (b x^2 - y^2 + 2xy) \) to be analytic are ___ .

• A. \( a = 1, b = -1 \)
• B. \( a = -1, b = 1 \)
• C. \( a = 1, b = 1 \)
• D. \( a = 1, b = 0 \)

Hint:

To check if a function is analytic, use the Cauchy-Riemann equations to determine if the function satisfies the necessary conditions.

Answer 1

The Correct Option is A

For the function to be analytic, the Cauchy-Riemann equations must be satisfied. These equations are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] where \( f(z) = u(x, y) + i v(x, y) \), and in this case: \[ u(x, y) = x^2 + a y^2 - 2xy \quad \text{and} \quad v(x, y) = b x^2 - y^2 + 2xy \] We compute the partial derivatives: \[ \frac{\partial u}{\partial x} = 2x - 2y, \quad \frac{\partial u}{\partial y} = 2a y - 2x \] \[ \frac{\partial v}{\partial x} = 2b x + 2y, \quad \frac{\partial v}{\partial y} = 2x - 2y \] Now, apply the Cauchy-Riemann equations: 1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \): \[ 2x - 2y = 2x - 2y \quad \text{(this is always true, so no condition on \( a \) and \( b \) from this equation)} \] 2. \( \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \): \[ 2a y - 2x = - (2b x + 2y) \] Simplifying: \[ 2a y - 2x = -2b x - 2y \] \[ 2a y + 2y = 2b x + 2x \] \[ y(2a + 2) = x(2b + 2) \] Dividing both sides by 2: \[ y(a + 1) = x(b + 1) \] For this equation to hold for all values of \( x \) and \( y \), we must have: \[ a + 1 = 0 \quad \text{and} \quad b + 1 = 0 \] Thus, \( a = -1 \) and \( b = -1 \). Therefore, the correct values of \( a \) and \( b \) that make the function analytic are \( a = 1 \) and \( b = -1 \).