Question

Let \( f: \mathbb{C} \to \mathbb{C} \) be defined by \( f(z) = |z|^2 - 5\bar{z} + 2 \).
Which one of the following is TRUE?

• A. \( f \) is differentiable at \( z = 5i \).
• B. \( f \) is differentiable at \( z = 5 \).
• C. \( f \) is differentiable at \( z = -5i \).
• D. \( f \) is differentiable at \( z = -5 \).

Hint:

Split f into real and imaginary parts and solve the Cauchy-Riemann equations for x and y.

Answer 1

The Correct Option is B

Step 1: Write f in terms of real and imaginary parts.
Let \( z = x + iy \) with \( x, y \) real. Then \( |z|^2 = x^2 + y^2 \) and \( \bar{z} = x - iy \).
\[ f(z) = |z|^2 - 5\bar{z} + 2 = (x^2 + y^2) - 5(x - iy) + 2 = (x^2 + y^2 - 5x + 2) + i(5y) \]
So if we write \( f = u + iv \), then \( u(x,y) = x^2 + y^2 - 5x + 2 \) and \( v(x,y) = 5y \).

Step 2: Recall the Cauchy-Riemann equations.
A function \( f = u + iv \) is complex differentiable at a point exactly where its partial derivatives are continuous there and satisfy the Cauchy-Riemann equations:
\[ u_x = v_y \quad \text{and} \quad u_y = -v_x \]
Since \( u \) and \( v \) here are polynomials in \( x, y \), their partials are continuous everywhere, so we only need to check where the two equations hold.

Step 3: Compute the partial derivatives and solve.
\[ u_x = 2x - 5, \quad u_y = 2y, \quad v_x = 0, \quad v_y = 5 \]
Setting \( u_x = v_y \) gives \( 2x - 5 = 5 \), so \( x = 5 \).
Setting \( u_y = -v_x \) gives \( 2y = 0 \), so \( y = 0 \).
Both Cauchy-Riemann equations hold only at the single point \( (x, y) = (5, 0) \), which is \( z = 5 \).

Step 4: Check each option against this point.
Option (A) asks about \( z = 5i \), that is \( (x,y) = (0,5) \): here \( x = 0 \neq 5 \), so the equations fail; not differentiable.
Option (C) asks about \( z = -5i \), that is \( (0,-5) \): again \( x = 0 \neq 5 \); not differentiable.
Option (D) asks about \( z = -5 \), that is \( (-5, 0) \): here \( x = -5 \neq 5 \); not differentiable.
Option (B) asks about \( z = 5 \), that is \( (5, 0) \): this is exactly the point we found, so \( f \) is differentiable there.

Final Answer:
\( f \) is complex differentiable only at the single point \( z = 5 \), matching option (B). \[ \boxed{z = 5} \]