Problem 39 - Advanced Engineering Mathematics

Question

We have $$\lim _{\Delta z \rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}.$$ If we let \(\Delta z \rightarrow 0\) along a horizontal line then \(\Delta z=\Delta x, \overline{\Delta z}=\Delta x,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1.$$ If we let \(\Delta z \rightarrow 0\) along a vertical line then \(\Delta z=i \Delta y, \overline{\Delta z}=-i \Delta y,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta y \rightarrow 0} \frac{-i \Delta y}{i \Delta y}=-1.$$ Since these two limits are not equal, \(f(z)=\bar{z}\) cannot be differentiable at any \(z\).

Step-by-Step Solution

Answer

\(f(z)=\bar{z}\) is not differentiable at any \(z\) since limits along different paths are not equal.

Step 1: Understanding the Limit Definition

The exercise involves evaluating a complex limit expression: \(\lim _{\Delta z \rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}\). This can be simplified to \(\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}\) by noting that \(\overline{z+\Delta z} - \bar{z}\) simplifies to \(\overline{\Delta z}\).

Step 2: Taking the Limit Along the Horizontal Line

When \(\Delta z\) approaches zero along a horizontal line, we let \(\Delta z = \Delta x\), thus \(\overline{\Delta z} = \Delta x\). The limit becomes \(\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}\), which simplifies to 1, since \(\Delta x/\Delta x\) is 1 for any nonzero \(\Delta x\).

Step 3: Taking the Limit Along the Vertical Line

When \(\Delta z\) approaches zero along a vertical line, we let \(\Delta z = i \Delta y\), hence \(\overline{\Delta z} = -i \Delta y\). The limit now becomes \(\lim _{\Delta y \rightarrow 0} \frac{-i \Delta y}{i \Delta y}\), which simplifies to -1 after canceling \(\Delta y\) and simplifying \(-i/i = -1\).

Step 4: Evaluating the Differentiability

To be differentiable, the limit must be the same regardless of the path of \(\Delta z\). However, the horizontal limit yields 1, and the vertical limit yields -1. Since these values are different, the function \(f(z) = \bar{z}\) is not differentiable at any \(z\).

Key Concepts

Complex Variables

Complex variables are an extension of regular variables into the realm of complex numbers. Unlike real numbers, complex numbers have a real part and an imaginary part, typically written as \( z = x + yi \), where \( x \) is the real part and \( y \) is the imaginary part.

• Real Part: \( ext{Re}(z) = x \)
• Imaginary Part: \( ext{Im}(z) = y \)

When dealing with complex functions, these components allow us to explore rich behaviors that aren't available in standard real number calculus.
To study these variables, it is crucial to master concepts like complex addition, multiplication, and especially conjugation, represented as \(\bar{z}\).
The conjugate of a complex number reverses the sign of the imaginary part: \( \bar{z} = x - yi \). This operation is particularly important for finding different types of limits and understanding differentiability in the complex domain.

Limit Definition

In calculus with complex numbers, limits function similarly to limits with real numbers. However, the two-dimensional nature of complex numbers makes these limits trickier. Consider the expression:\[\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}\] Finding this limit requires evaluating it as \( \Delta z \) approaches zero via different paths, such as horizontally or vertically in the complex plane.
When attempting these evaluations, note:

• Horizontal paths involve real changes, setting \( \Delta z = \Delta x \), leading to simplifications like \( \frac{\Delta x}{\Delta x} = 1 \).
• Vertical paths involve changes along the imaginary axis, \( \Delta z = i \Delta y \), giving simplifications like \( \frac{-i \Delta y}{i \Delta y} = -1 \).

The behavior of these limits across paths helps determine the differentiability of complex functions.

Differentiability

For a function of a complex variable to be differentiable at a point, the limit of the difference quotient must exist and be the same, irrespective of the path used to approach zero. The limit in question:\[\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}\]reveals path-dependent values—specifically, it reaches 1 along horizontal paths and -1 along vertical paths.
Differentiability in the complex sense implies a condition stronger than in real function scenarios. The disparity between limit values indicates that \( f(z) = \bar{z} \) is not complex differentiable at any point.
Check for differentiability using the Cauchy-Riemann equations, another method to ascertain if a complex function maintains the property at a given point. This particular function fails that test, too, echoing conclusions from the limit approach.

Path Dependence

Path dependence is a crucial concept in evaluating the differentiability of complex functions. It refers to situations where the outcome of a limit differs based on how zero is approached in the complex plane.
For example, with the function \( f(z) = \bar{z} \), the path-dependence of the limit:\[ \lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z} \] demonstrates the deep link between the path and the final outcome.

• The horizontal approach gives \( 1 \).
• The vertical approach results in \( -1 \).

Such differences consign \( f(z) = \bar{z} \) as non-differentiable. When solving complex problems, always consider path dependence. This principle ensures that solutions are correct across all potential trajectories in the complex plane.