Problem 1 - Advanced Engineering Mathematics

Question

In Problems 1 and 2 , show that \(z=0\) is a removable singularity of the given function. Supply a definition of \(f(0)\) so that \(f\) is analytic at \(z=0\). $$ f(z)=\frac{e^{2 z}-1}{z} $$

Step-by-Step Solution

Answer

Define \( f(0) = 2 \) to make \( z=0 \) a removable singularity and \( f \) analytic.

Step 1: Recognize the problem type

We need to determine if the point \( z=0 \) is a removable singularity for the function \( f(z)=\frac{e^{2z}-1}{z} \). A removable singularity exists if the limit exists as \( z \to 0 \).

Step 2: Limit calculation

To find if there's a removable singularity at \( z=0 \), we compute the limit \( \lim_{z \to 0} \frac{e^{2z}-1}{z}\). This limit will tell us if the function can be extended to be analytic at \( z=0 \).

Step 3: Apply L'Hôpital's rule

Notice that both the numerator \( e^{2z}-1 \) and denominator \( z \) approach 0 as \( z \to 0 \). Apply L'Hôpital's rule, which is suitable because this is a 0/0 indeterminate form. Differentiate the numerator and denominator: the derivative of \( e^{2z} - 1 \) is \( 2e^{2z} \), and the derivative of \( z \) is \( 1 \).

Step 4: Evaluate the new limit

Substitute the derivatives into the limit: \( \lim_{z \to 0} \frac{2e^{2z}}{1} = \lim_{z \to 0} 2e^{2z} \). Evaluate this limit, which simplifies to \( 2 \cdot 1 = 2 \).

Step 5: Define \( f(0) \)

Since the limit \( \lim_{z \to 0} \frac{e^{2z}-1}{z} \) exists and equals 2, define \( f(0) = 2 \). This definition removes the singularity and makes \( f \) analytic at \( z=0 \).

Key Concepts

Analytic Function

An analytic function, also referred to as a holomorphic function, is a complex function that is differentiable at every point in its domain. This means the function has a derivative at each point, which implies it is smooth and has no sharp corners or discontinuities.
Unlike real-valued functions, analytic functions of a complex variable have some very special properties:

• They are infinitely differentiable, meaning you can find derivatives of all orders.
• Their behavior around a small region is completely determined by the values and derivatives at a single point.
• An analytic function can be expressed as a power series.

For a function to be analytic at a certain point, such as in our exercise, any singularities—points where the function might not be well-defined—must be addressed. If a function is represented by a limit at a singular point, as shown in our original exercise, it can often be made analytic by properly defining its value at that point. This effectively "removes" the singularity by making the function continuous and differentiable there.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate the limits of indeterminate forms. Indeterminate forms often arise as limits that initially appear ambiguous, such as 0/0 or ∞/∞. L'Hôpital's Rule provides us with a method to resolve these forms by finding derivatives.
The rule states that if the limit of a function as variable approaches a given point results in 0/0 or ∞/∞, you can take the derivatives of the numerator and denominator separately and then evaluate the limit again. More specifically, if \[\lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \frac{0}{0} \text{ or } \frac{\infty}{\infty},\]then\[\lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to a}} \frac{{f'(x)}}{{g'(x)}}.\]This rule was essential in the original exercise to calculate the limit and determine the value at which the function is continuous and differentiable at specific points. Without L'Hôpital's Rule, resolving such indeterminate forms would be much more challenging, especially in the realm of complex functions.

Limit Calculation

Understanding limit calculation is fundamental in pinpointing how a function behaves as the variable approaches a particular point. In complex analysis, calculating the limit at specific points is crucial, especially when dealing with singularities.
The process often involves:

• Identifying if the original function leads to an indeterminate form.
• Determining methods like L'Hôpital's Rule to resolve such forms and find numeric limits.
• Interpreting the results to address continuity and differentiability issues.

In the given exercise, we needed to calculate the limit of \(\lim_{{z \to 0}} \frac{{e^{2z}-1}}{z}\) to confirm if a removable singularity existed at \(z=0\). Calculating this limit correctly allows us to redefine the function at that point, making it continuous and analytic. Successfully executing limit calculations ensures a deeper understanding of function behavior and their possible extensions, ensuring complete analytic character in the domain.