Problem 4
Question
$$u=y, \quad v=x ; \quad \frac{\partial u}{\partial x}=0=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=1,-\frac{\partial v}{\partial x}=-1 . \text { since } 1 \neq-1, f \text { is not analytic at any point. }$$
Step-by-Step Solution
Verified Answer
The function is not analytic because the Cauchy-Riemann conditions are not met.
1Step 1: Understanding the Given Variables
We are provided with two variables: \( u = y \) and \( v = x \). These form the real (\( u \)) and imaginary (\( v \)) parts of a potential complex function \( f(z) = u + iv \).
2Step 2: Calculating Partial Derivatives
Compute the partial derivatives: \( \frac{\partial u}{\partial x} = 0 \), \( \frac{\partial v}{\partial y} = 0 \), \( \frac{\partial u}{\partial y} = 1 \), and \( -\frac{\partial v}{\partial x} = -1 \). These derivatives will be crucial for checking the Cauchy-Riemann equations.
3Step 3: Checking Cauchy-Riemann Equations
For \( f \) to be analytic, Cauchy-Riemann equations need to hold: \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \) and \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \). We have \( 0 = 0 \) for the first equation. However, for the second, \( 1 eq -1 \), which means this condition does not hold.
4Step 4: Conclusion: Analyticity
Since the Cauchy-Riemann equations are not satisfied, the function \( f(z) = u + iv \) is not analytic at any point.
Key Concepts
Cauchy-Riemann EquationsAnalytic FunctionsPartial Derivatives
Cauchy-Riemann Equations
In complex analysis, the Cauchy-Riemann equations are fundamental because they determine whether a function is analytic. This means the function is differentiable at every point in its domain and has a complex derivative. For a function \( f(z) = u(x, y) + iv(x, y) \), where \( u \) and \( v \) are real-valued functions of two real variables \( x \) and \( y \), the Cauchy-Riemann equations are:
In the context of our exercise, both equations were checked but only the first equation held true. It was shown that \( \frac{\partial u}{\partial y} = 1 \) was not equal to \( -\frac{\partial v}{\partial x} = -1 \), violating the second Cauchy-Riemann equation. As a result, the function \( f \) is not analytic.
- \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
- \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \)
In the context of our exercise, both equations were checked but only the first equation held true. It was shown that \( \frac{\partial u}{\partial y} = 1 \) was not equal to \( -\frac{\partial v}{\partial x} = -1 \), violating the second Cauchy-Riemann equation. As a result, the function \( f \) is not analytic.
Analytic Functions
Analytic functions are a special class of functions in complex analysis that are not only differentiable but also possess derivatives of all orders. Their beauty lies in their smoothness and the fact that they can be expressed as power series. Notably, these functions are typically referred to as holomorphic, especially when defined in a domain.
For a function \( f(z) \) to be analytic at a point \( z_0 \), it must satisfy several criteria:
For a function \( f(z) \) to be analytic at a point \( z_0 \), it must satisfy several criteria:
- Existence of a derivative at every point close to \( z_0 \)
- Satisfaction of the Cauchy-Riemann equations in a neighborhood around \( z_0 \)
- Continuous partial derivatives
Partial Derivatives
Partial derivatives are essential tools in mathematics, particularly in complex analysis. They help us understand how a function changes when only one of its input variables changes while keeping everything else constant. This is especially useful in multivariable functions like those seen in complex analysis where both real and imaginary parts are functions of two real variables \( x \) and \( y \).
In our exercise, the partial derivatives of the real part \( u \) with respect to \( x \), and \( y \), and the partial derivatives of the imaginary part \( v \) with respect to \( x \), and \( y \) were calculated. These derivatives enabled us to check the Cauchy-Riemann equations to determine the analytic nature of the function. Understanding partial derivatives is crucial because they give insight into the function's behavior and are fundamental in verifying whether the essential conditions for analyticity are satisfied.
In our exercise, the partial derivatives of the real part \( u \) with respect to \( x \), and \( y \), and the partial derivatives of the imaginary part \( v \) with respect to \( x \), and \( y \) were calculated. These derivatives enabled us to check the Cauchy-Riemann equations to determine the analytic nature of the function. Understanding partial derivatives is crucial because they give insight into the function's behavior and are fundamental in verifying whether the essential conditions for analyticity are satisfied.
Other exercises in this chapter
Problem 4
$$\cos (2-4 i)=\cos (2) \cosh (-4)-\sin (2) \sinh (-4)=-11.3642-24.8147 i$$
View solution Problem 4
$$e^{2-\frac{\pi}{2} i}=e^{2} \cos \left(-\frac{\pi}{2}\right)+i e^{2} \sin \left(-\frac{\pi}{2}\right)=-e^{2} i$$
View solution Problem 4
$$6\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$$
View solution Problem 4
$$i^{11}=i\left(i^{2}\right)^{5}=i(-1)^{5}=-i$$
View solution