To prove that the derivative $\frac{d}{dz} (z^2 \overline{z})$ does not exist anywhere, we need to examine the function $f(z) = z^2 \overline{z}$ in the context of complex differentiability.

1. Understanding $f(z) = z^2 \overline{z}$

In complex analysis, a function $f(z)$ is differentiable at a point $z_0$ if the limit defining the derivative, $f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h},$ exists. This requires $f(z)$ to satisfy the Cauchy-Riemann equations at $z_0$.

Cauchy-Riemann Equations

If $f(z) = u(x, y) + iv(x, y)$ with $z = x + iy$, the Cauchy-Riemann equations are: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$ These equations are necessary for $f(z)$ to be complex differentiable at any point in the domain.

2. Expressing $f(z) = z^2 \overline{z}$ in Terms of $x$ and $y$

Let's rewrite $z^2 \overline{z}$ in terms of real and imaginary parts:

• Let $z = x + iy$, so $\overline{z} = x - iy$.
• Therefore, $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$.
• Consequently, $z^2 \overline{z} = (x^2 - y^2 + 2ixy)(x - iy).$ Expanding this, $z^2 \overline{z} = (x^2 - y^2)x + 2ix^2y - (x^2 - y^2)iy - 2y^2x.$

Breaking it down into real and imaginary parts, we get: $u(x, y) = x^3 - 3xy^2 \quad \text{and} \quad v(x, y) = y(x^2 - y^2).$

3. Calculating Partial Derivatives and Checking the Cauchy-Riemann Equations

Now, calculate the partial derivatives: $\frac{\partial u}{\partial x} = 3x^2 - 3y^2, \quad \frac{\partial u}{\partial y} = -6xy,$ $\frac{\partial v}{\partial x} = 2xy, \quad \frac{\partial v}{\partial y} = x^2 - 3y^2.$

To satisfy the Cauchy-Riemann equations, the following must hold: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \Rightarrow \quad 3x^2 - 3y^2 = x^2 - 3y^2.$ Simplifying, this gives: $2x^2 = 0 \quad \Rightarrow \quad x = 0.$

For the second equation: $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \quad \Rightarrow \quad -6xy = -2xy.$ Simplifying, this gives: $-4xy = 0 \quad \Rightarrow \quad x = 0 \text{ or } y = 0.$

4. Conclusion

The solutions to the Cauchy-Riemann equations are restricted to $x = 0$ and $y = 0$, which is only at the origin $(0,0)$. However, even at the origin, the partial derivatives are not continuous, as seen from the differences in terms involving $x$ and $y$.

Since the Cauchy-Riemann equations are not satisfied anywhere except trivially at isolated points, $f(z) = z^2 \overline{z}$ is nowhere complex differentiable.

Thus, the derivative $\frac{d}{dz} (z^2 \overline{z})$ does not exist anywhere.

Would you like more details on any part of this proof, or have any questions?

---

Here are some questions to consider:

1. What are the implications of a function failing the Cauchy-Riemann equations everywhere?
2. How does the presence of $\overline{z}$ in a function impact its differentiability?
3. Could this approach apply to other functions involving $z$ and $\overline{z}$?
4. Why are the Cauchy-Riemann equations necessary for complex differentiability?
5. How do real and imaginary parts interplay in complex differentiable functions?

Tip: In complex analysis, when examining differentiability, always check for the Cauchy-Riemann equations. They are essential criteria for determining if a complex function has a complex derivative at a point.