Let's solve the problem step-by-step: --- ### Step 1: Use the Cauchy-Riemann equations to determine where \( f \) is differentiable. Given: \[ f(z) = \bar{z} + z^2 \] Let \( z = x + iy \), so \( \bar{z} = x - iy \). Expand \( z^2 \): \[ z^2 = (x + iy)^2 = x^2 + 2ixy - y^2 \] So, \[ f(z) = (x - iy) + (x^2 - y^2 + 2ixy) = (x + x^2 - y^2) + i(2xy - y) \] Let \( u(x, y) = x + x^2 - y^2 \) and \( v(x, y) = 2xy - y \). The **Cauchy-Riemann equations** are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Compute the partial derivatives: - \( \frac{\partial u}{\partial x} = 1 + 2x \) - \( \frac{\partial u}{\partial y} = -2y \) - \( \frac{\partial v}{\partial x} = 2y \) - \( \frac{\partial v}{\partial y} = 2x - 1 \) Plug into the Cauchy-Riemann equations: 1. \( 1 + 2x = 2x - 1 \implies 1 = -1 \) (**Impossible**) 2. \( -2y = -2y \) (**Always true**) **Step 1 Explanation:** Since the first equation leads to a contradiction, the Cauchy-Riemann equations are never satisfied at any point in the complex plane. --- ### Step 2: Is \( f \) analytic anywhere? A function \( f \) is **analytic** at a point if it is differentiable in an open neighborhood around that point. From Step 1, \( f \) is **nowhere differentiable**, so it cannot be analytic anywhere. **Step 2 Explanation:** Since \( f \) does not satisfy the Cauchy-Riemann equations at any point, it is **nowhere analytic**. --- ## **Final Answers** **(a)** \( f(z) \) is nowhere differentiable because the Cauchy-Riemann equations are never satisfied. **(b)** \( f(z) \) is nowhere analytic, since it is nowhere differentiable.

Let's analyze the problem step-by-step: --- ### Step 1: Use the Cauchy-Riemann equations to determine where \( f(z) \) is differentiable. **Given function:** \[ f(z) = \bar{z} + z^2 \] Expressing in terms of \( x \) and \( y \), where \( z = x + iy \): - \( \bar{z} = x - iy \) - \( z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy \) Thus, \[ f(z) = (x - iy) + (x^2 - y^2 + 2ixy) = (x + x^2 - y^2) + i(2xy - y) \] Define: \[ u(x, y) = x + x^2 - y^2 \] \[ v(x, y) = 2xy - y \] The **Cauchy-Riemann equations** are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Compute derivatives: - \( u_x = 1 + 2x \) - \( u_y = -2y \) - \( v_x = 2y \) - \( v_y = 2x - 1 \) Set up the equations: 1. \( 1 + 2x = 2x - 1 \) 2. \( -2y = -2y \) **Solve equation 1:** \[ 1 + 2x = 2x - 1 \implies 1 = -1 \] which is **impossible**. **Conclusion for Step 1:** Since the first Cauchy-Riemann equation cannot be satisfied at any point (no solutions), **\(f(z)\) is nowhere differentiable**. --- ### Step 2: Is \( f(z) \) analytic anywhere? A function is **analytic** at a point if it is complex differentiable in some neighborhood of that point. Since \(f(z)\) is **not differentiable anywhere** (from Step 1), it cannot be **analytic anywhere**. **Final conclusion:** \[ \boxed{ \text{(a)} \quad \text{f is nowhere differentiable} \\ \text{(b)} \quad \text{f is nowhere analytic} } \]