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To determine which function \( f(z) \) satisfies the Cauchy-Riemann equations throughout the complex plane, let's analyze each function.

The Cauchy-Riemann equations are:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x},
\]
where \( f(z) = u(x, y) + iv(x, y) \) with \( u(x, y) \) as the real part and \( v(x, y) \) as the imaginary part.

Let's examine each option:

1. **\( f(z) = x^2 - iy^2 \):**
   - \( u(x, y) = x^2 \), \( v(x, y) = -y^2 \).
   - Calculating partial derivatives:
     \[
     \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = -2y.
     \]
     Since \( \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} \), this function does **not** satisfy the Cauchy-Riemann equations.

2. **\( f(z) = (x^2 - y^2) - i \cdot 2xy \):**
   - \( u(x, y) = x^2 - y^2 \), \( v(x, y) = -2xy \).
   - Calculating partial derivatives:
     \[
     \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = -2x,
     \]
     \[
     \frac{\partial u}{\partial y} = -2y, \quad \frac{\partial v}{\partial x} = -2y.
     \]
     Both Cauchy-Riemann equations are satisfied:
     \[
     \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.
     \]
     This function **does** satisfy the Cauchy-Riemann equations throughout the complex plane.

3. **\( f(z) = (x^2 - y^2) + i \cdot 2xy \):**
   - \( u(x, y) = x^2 - y^2 \), \( v(x, y) = 2xy \).
   - Calculating partial derivatives:
     \[
     \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 2x,
     \]
     \[
     \frac{\partial u}{\partial y} = -2y, \quad \frac{\partial v}{\partial x} = 2y.
     \]
     Here, \( \frac{\partial u}{\partial y} \neq -\frac{\partial v}{\partial x} \), so this function does **not** satisfy the Cauchy-Riemann equations.

4. **\( f(z) = x^2 + iy^2 \):**
   - \( u(x, y) = x^2 \), \( v(x, y) = y^2 \).
   - Calculating partial derivatives:
     \[
     \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 2y.
     \]
     Since \( \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} \), this function does **not** satisfy the Cauchy-Riemann equations.

### Conclusion
The function \( f(z) = (x^2 - y^2) - i \cdot 2xy \) is the only one that satisfies the Cauchy-Riemann equations throughout the complex plane.

Would you like more details on the calculations? 

Here are 5 related questions for further study:
1. What are the Cauchy-Riemann equations, and why are they important in complex analysis?
2. How do we interpret a function satisfying the Cauchy-Riemann equations geometrically?
3. Can a function satisfy the Cauchy-Riemann equations at a point but not be differentiable there?
4. What are the conditions for a complex function to be analytic?
5. How are harmonic functions related to complex differentiable functions?

**Tip:** To confirm that a complex function is differentiable, check that it satisfies the Cauchy-Riemann equations and that the partial derivatives are continuous.

Let z = x + iy. Which one of the following functions satisfy Cauchy-Riemann equations throughout the complex plane?

Explore how to determine which functions satisfy the Cauchy-Riemann equations across the complex plane. This problem delves into the fundamentals of complex analysis, involving partial derivatives and conditions for differentiability in complex functions.

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