Consider the following functions:(a) f(z) = x3(1 + i) − y3(1 − i)x2 + y2, (z ̸= 0), f(0) = 0,

To analyze the function $f(z) = x^3(1 + i) - y^3(1 - i)x^2 + y^2$ where $z = x + iy$, we will look at this function in terms of the complex variables and see if it has any specific properties such as continuity, differentiability, or any interesting features.

1. Break Down the Problem

   • The function is given in terms of its real and imaginary components.
   • Identify the real part $u(x, y)$ and the imaginary part $v(x, y)$ of the function $f(z)$.

2. Relevant Concepts

   • A function of a complex variable is analytic if it satisfies the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

3. Analysis and Detail

   • First, we will express $f(z)$ in terms of $u(x, y)$ and $v(x, y)$: $f(z) = u(x, y) + iv(x, y)$
   • Here, $u(x, y) = x^3(1 + i) - y^3(1 - i)x^2 + y^2 = x^3 + iy^3 - y^3x^2 + y^2$
   • Considering $u(x, y)$ and isolating the real and imaginary parts:
     • Real part $u(x, y) = x^3 - y^3x^2 + y^2$
     • Imaginary part $v(x, y) = y^3 + x^3$

4. Verify and Summarize

   • Now, we calculate the partial derivatives:

     • $\frac{\partial u}{\partial x} = 3x^2 - 2y^3$
     • $\frac{\partial u}{\partial y} = -3y^2x + 2y$
     • $\frac{\partial v}{\partial x} = 3x^2$
     • $\frac{\partial v}{\partial y} = 3y^2$

   • Check the Cauchy-Riemann equations:

     • $\frac{\partial u}{\partial x} = 3x^2 - 2y^3$ is not equal to $\frac{\partial v}{\partial y} = 3y^2$ in general.
     • $\frac{\partial u}{\partial y} = -3y^2x + 2y$ is not equal to $-\frac{\partial v}{\partial x} = -3x^2$.

Thus, since the Cauchy-Riemann equations do not hold, $f(z)$ is not differentiable in the complex sense in any area where $z \neq 0$.

Final Answer

The function $f(z)$ is not analytic anywhere in $\mathbb{C} \setminus \{0\}$.