To determine which function satisfies the Cauchy-Riemann equations throughout the complex plane, let’s analyze each function one by one.

Given:

• $z = x + iy$
• $f(z) = u(x, y) + iv(x, y)$, where $u$ is the real part and $v$ is the imaginary part.

The Cauchy-Riemann equations state that for $f(z)$ to be differentiable, the following must hold: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

1. $f(z) = (x^2 - y^2) - i2xy$

• Here, $u(x, y) = x^2 - y^2$ and $v(x, y) = -2xy$.
• Compute the partial derivatives: $\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y$ $\frac{\partial v}{\partial x} = -2y, \quad \frac{\partial v}{\partial y} = -2x$
• Since $\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} \neq -\frac{\partial v}{\partial x}$, the Cauchy-Riemann equations are not satisfied.

2. $f(z) = x^2 - iy^2$

• Here, $u(x, y) = x^2$ and $v(x, y) = -y^2$.
• Compute the partial derivatives: $\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = 0$ $\frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = -2y$
• Since $\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} \neq -\frac{\partial v}{\partial x}$, the Cauchy-Riemann equations are not satisfied.

3. $f(z) = (x^2 - y^2) + i2xy$

• Here, $u(x, y) = x^2 - y^2$ and $v(x, y) = 2xy$.
• Compute the partial derivatives: $\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = -2y$ $\frac{\partial v}{\partial x} = 2y, \quad \frac{\partial v}{\partial y} = 2x$
• Since $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$, the Cauchy-Riemann equations are satisfied.

4. $f(z) = x^2 + iy^2$

• Here, $u(x, y) = x^2$ and $v(x, y) = y^2$.
• Compute the partial derivatives: $\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = 0$ $\frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = 2y$
• Since $\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} \neq -\frac{\partial v}{\partial x}$, the Cauchy-Riemann equations are not satisfied.

Conclusion

The only function that satisfies the Cauchy-Riemann equations throughout the complex plane is: $f(z) = (x^2 - y^2) + i2xy$

Let me know if you want more details on any part of the solution or if you have further questions.

Here are 5 related questions to deepen your understanding:

1. How are the Cauchy-Riemann equations derived from complex differentiability?
2. Can a function satisfy the Cauchy-Riemann equations at a point but not be analytic?
3. How does differentiability in complex analysis differ from real analysis?
4. What is the significance of harmonic functions in relation to the Cauchy-Riemann equations?
5. How do Cauchy-Riemann equations connect to conformal mappings?

Tip: The Cauchy-Riemann equations provide a necessary condition for a function to be analytic, but satisfying them alone doesn’t guarantee analyticity unless the function is also continuous.