State and derive Cauchy-Riemann equations for analytic functions. विश्लेघणात्मक कार्यों के लिए कॉची-रीमैन समीकरण बताइए और व्युत्पन्न कीजिए। Find the analytic function whose real part is $$ x^{2}-y^{2} $$. वह विश्लेषणात्मक फलन ज्ञात कीजिए जिसका वास्तािक भाग $$ x^{2}-y^{2} $$ है। Let $$ f z) $$ be an analytic function of $$ z $$ in a region D of the $$ z $$-plane and let $$ f(z) $$ may be zero inside $$ D $$. Then prove that the mapping $$ w=f(z) $$ may not be conformal at the points of $$ D $$.

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### 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 $$ f(z) = 0 $$.
**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 $$ f(z) = 0 $$.

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