Question

The function \( f(z) = |z|^2 \) is differentiable, at

• A. \( z = 0 \)
• B. for all \( z \in \mathbb{C} \)
• C. no \( z \in \mathbb{C} \)
• D. \( z \neq 0 \)

Hint:

For questions about differentiability of complex functions involving \(|z|\), \(\bar{z}\), Re(z), or Im(z), always revert to the Cauchy-Riemann equations in Cartesian form (\(u_x = v_y, u_y = -v_x\)). These functions are typically not differentiable anywhere or are differentiable only at specific points like \(z=0\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A complex function \( f(z) \) is differentiable at a point \( z_0 \) if it satisfies the Cauchy-Riemann equations at that point and its first-order partial derivatives are continuous there. Let \( z = x + iy \), so \( f(z) = u(x,y) + iv(x,y) \). The Cauchy-Riemann equations are \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \) and \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \).

Step 2: Key Formula or Approach:
First, express the function \( f(z) = |z|^2 \) in terms of its real and imaginary parts. \[ f(z) = |x+iy|^2 = (\sqrt{x^2+y^2})^2 = x^2+y^2 \] So, the real part is \( u(x,y) = x^2+y^2 \) and the imaginary part is \( v(x,y) = 0 \).

Step 3: Detailed Explanation:
Now, we compute the first-order partial derivatives: \[ \frac{\partial u}{\partial x} = 2x \] \[ \frac{\partial u}{\partial y} = 2y \] \[ \frac{\partial v}{\partial x} = 0 \] \[ \frac{\partial v}{\partial y} = 0 \] For the Cauchy-Riemann equations to hold, we need: 1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \implies 2x = 0 \implies x = 0 \)
2. \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \implies 2y = -0 \implies y = 0 \)
Both equations are satisfied only when \( x=0 \) and \( y=0 \). This corresponds to the point \( z = 0 + i0 = 0 \).
The partial derivatives are polynomials in \(x\) and \(y\), so they are continuous everywhere. Since the Cauchy-Riemann equations are satisfied only at \(z=0\) and the partials are continuous, the function \(f(z) = |z|^2\) is differentiable only at \(z=0\).

Step 4: Final Answer:
The function is differentiable only at \( z = 0 \).