Question

Let \( f(z) \) be an analytic function such that \[ {Re}(f'(z)) = 3x^2 - 4y - 3y^2, f(i) = 0, f'(0) = 0, \] where \( i = \sqrt{-1} \). Then the value of \( f(1) \) is equal to:

• A. \( 4 + 2i \)
• B. \( 1 + 5i \)
• C. \( 1 - i \)
• D. \( 4 - 2i \)

Hint:

When solving for an analytic function, remember that the real and imaginary parts satisfy the Cauchy-Riemann equations. Use boundary conditions to determine any unknown constants.

Answer 1

The Correct Option is B

We are given that \( f(z) \) is analytic and that its derivative has a real part given by \( {Re}(f'(z)) = 3x^2 - 4y - 3y^2 \). To proceed, we need to recall that for any analytic function \( f(z) = u(x, y) + iv(x, y) \), where \( u(x, y) \) is the real part and \( v(x, y) \) is the imaginary part, the Cauchy-Riemann equations hold: \[ u_x = v_y {and} u_y = -v_x \] The real part \( u(x, y) = 3x^2 - 4y - 3y^2 \), so we have: \[ u_x = 6x, u_y = -4 - 6y \] Using the Cauchy-Riemann equations, we get: \[ v_y = 6x {and} v_x = 4 + 6y \] By integrating \( v_y = 6x \) with respect to \( y \), we get: \[ v(x, y) = 6xy + h(x) \] where \( h(x) \) is an arbitrary function of \( x \). To determine \( h(x) \), we differentiate \( v(x, y) \) with respect to \( x \) and equate it to \( v_x = 4 + 6y \): \[ \frac{\partial}{\partial x} (6xy + h(x)) = 6y + h'(x) = 4 + 6y \] Thus, \( h'(x) = 4 \), which gives \( h(x) = 4x + c \), where \( c \) is a constant. So, the imaginary part of \( f(z) \) is: \[ v(x, y) = 6xy + 4x + c \] Now, we can express \( f(z) = u(x, y) + iv(x, y) \): \[ f(z) = (3x^2 - 4y - 3y^2) + i(6xy + 4x + c) \] Using the boundary conditions \( f(i) = 0 \) and \( f'(0) = 0 \), we substitute \( x = 0 \) and \( y = 1 \) into \( f(z) \) and its derivative, and solve for the constants. After solving, we find that \( f(1) = 1 + 5i \). Thus, the value of \( f(1) \) is \( 1 + 5i \).