Question

The complex-valued function \( f(z) = iz - |z|^2 \) is analytic at ...........

• A. nowhere
• B. everywhere
• C. \( z = 1 \)
• D. \( z = 0 \)

Hint:

A complex function \( f(z) = u + iv \) is analytic at a point if the Cauchy-Riemann equations hold and the partial derivatives of \( u \) and \( v \) are continuous there.

Answer 1

The Correct Option is D

Step 1: Let \( z = x + iy \). Then \[ f(z) = i(x + iy) - |z|^2 = i(x + iy) - (x^2 + y^2) = ix - y - (x^2 + y^2) \] So, writing in terms of real and imaginary parts: \[ f(z) = ( -x^2 - y^2 - y ) + i x ⇒ u(x, y) = -x^2 - y^2 - y,\quad v(x, y) = x \] Step 2: Check the Cauchy-Riemann (CR) equations:
Cauchy-Riemann conditions: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Compute derivatives: \[ \frac{\partial u}{\partial x} = -2x,\quad \frac{\partial u}{\partial y} = -2y - 1,\quad \frac{\partial v}{\partial x} = 1,\quad \frac{\partial v}{\partial y} = 0 \] At \( z = 0 ⇒ x = 0,\ y = 0 \):
\[ \frac{\partial u}{\partial x} = 0 = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -1 = -\frac{\partial v}{\partial x} \] So the CR equations are satisfied at \( z = 0 \) only.
Step 3: Function is analytic only at points where CR equations hold and partial derivatives are continuous. 
All functions are polynomials (smooth), so partials are continuous.
Therefore, \( f(z) \) is analytic only at \( \boxed{z = 0} \).