Answer
**Cauchy-Riemann Equations:**
For an analytic function \( f(z) = u(x, y) + iv(x, y) \), the Cauchy-Riemann equations are:
1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
2. \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)
**Finding the Analytic Function:**
Given \( u(x, y) = x^2 - y^2 \), the imaginary part \( v(x, y) \) is \( 2xy + C \), where \( C \) is a constant. Therefore, the analytic function is:
\[
f(z) = x^2 - y^2 + i(2xy + C)
\]
**Non-Conformality of Mapping \( w = f(z) \):**
If \( f(z) \) is zero at a point \( z_0 \) in region \( D \), then \( f'(z_0) = 0 \). This means the mapping \( w = f(z) \) is not conformal at \( z_0 \) because angles and shapes are not preserved at that point.
**Summary:**
- **Cauchy-Riemann Equations:** Ensure that the real and imaginary parts of an analytic function satisfy these partial differential equations.
- **Analytic Function:** \( f(z) = x^2 - y^2 + i(2xy + C) \)
- **Mapping Behavior:** The mapping \( w = f(z) \) is not conformal at points where \( f(z) = 0 \).
Solution
### Cauchy-Riemann Equations
The Cauchy-Riemann equations are a set of two partial differential equations that are satisfied by the real and imaginary parts of an analytic function. If \( f(z) = u(x, y) + iv(x, y) \), where \( z = x + iy \), then the Cauchy-Riemann equations state that:
1. \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
2. \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)
#### Derivation of Cauchy-Riemann Equations
To derive the Cauchy-Riemann equations, we start with the definition of an analytic function. A function \( f(z) \) is analytic in a region if it is differentiable at every point in that region. The derivative of \( f(z) \) can be expressed as:
\[
f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
\]
Let \( \Delta z = \Delta x + i \Delta y \). Then, we can express \( f(z + \Delta z) \) as:
\[
f(z + \Delta z) = f(x + \Delta x + i(y + \Delta y)) = u(x + \Delta x, y + \Delta y) + iv(x + \Delta x, y + \Delta y)
\]
Using Taylor expansion, we can write:
\[
u(x + \Delta x, y + \Delta y) \approx u(x, y) + \frac{\partial u}{\partial x} \Delta x + \frac{\partial u}{\partial y} \Delta y
\]
\[
v(x + \Delta x, y + \Delta y) \approx v(x, y) + \frac{\partial v}{\partial x} \Delta x + \frac{\partial v}{\partial y} \Delta y
\]
Substituting these into the derivative expression and separating real and imaginary parts leads to the Cauchy-Riemann equations.
### Finding the Analytic Function
Given that the real part of the analytic function \( f(z) \) is \( u(x, y) = x^2 - y^2 \), we need to find the imaginary part \( v(x, y) \).
Using the Cauchy-Riemann equations:
1. From \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \):
\[
\frac{\partial u}{\partial x} = 2x \implies \frac{\partial v}{\partial y} = 2x
\]
2. From \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \):
\[
\frac{\partial u}{\partial y} = -2y \implies -\frac{\partial v}{\partial x} = -2y \implies \frac{\partial v}{\partial x} = 2y
\]
Now we have the system of equations:
\[
\frac{\partial v}{\partial y} = 2x \quad (1)
\]
\[
\frac{\partial v}{\partial x} = 2y \quad (2)
\]
Integrating equation (1) with respect to \( y \):
\[
v(x, y) = 2xy + g(x)
\]
where \( g(x) \) is an arbitrary function of \( x \).
Now, substituting \( v(x, y) \) into equation (2):
\[
\frac{\partial}{\partial x}(2xy + g(x)) = 2y + g'(x) = 2y
\]
This implies \( g'(x) = 0 \), so \( g(x) \) is a constant. Let \( g(x) = C \).
Thus, the imaginary part is:
\[
v(x, y) = 2xy + C
\]
Therefore, the analytic function is:
\[
f(z) = (x^2 - y^2) + i(2xy + C)
\]
### Proving Non-Conformality of Mapping \( w = f(z) \)
Let \( f(z) \) be an analytic function that may be zero inside a region \( D \). If \( f(z_0) = 0 \) for some point \( z_0 \in D \), then the derivative \( f'(z_0) \) must also be zero (by the property of analytic functions).
The mapping \( w = f(z) \) is conformal if the derivative \( f'(z) \) is non-zero. At points where \( f(z) = 0 \), we have:
\[
f'(z_0) = 0
\]
This means that the mapping is not conformal at \( z_0 \) because the angle and shape of small figures around \( z_0 \) are not preserved. Thus, the mapping \( w = f(z) \) may not be conformal at the points where \( f(z) = 0 \).
### Conclusion
1. The Cauchy-Riemann equations are derived and stated.
2. The analytic function whose real part is \( x^2 - y^2 \) is \( f(z) = (x^2 - y^2) + i(2xy + C) \).
3. The mapping \( w = f(z) \) is not conformal at points where \( f(z) = 0 \).
Reviewed and approved by the UpStudy tutoring team
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