Question

Show that for \(g\) and \(h\) analytic functions at \(z_{0}\), with \(g\left(z_{0}\right) \neq 0, h\left(z_{0}\right)=0\), and \(h^{\prime}\left(z_{0}\right) \neq 0\) $$ \operatorname{Res}\left[\frac{g(z)}{h(z)} ; z_{0}\right]=\frac{g\left(z_{0}\right)}{h^{\prime}\left(z_{0}\right)} $$

Short answer

Under the given conditions where \(h(z_0) = 0\), \(h'(z_0) ≠ 0\) and \(g(z_0) ≠ 0\), it's been demonstrated that the residue of a function \(f(z) = g(z)/h(z)\) at a point \(z_0\) is given by \(Res[f(z); z_0] = g(z_0)/h'(z_0)\).

Step-by-step solution

Identify the functions and their properties

Firstly, identify that \(g\) and \(h\) are analytic functions at \(z_0\) which means that they are complex differentiable at point \(z_0\). Additionally, \(g(z_0)\) is not equal to zero and \(h(z_0)\) is equal to zero. The derivative of \(h\) at \(z_0\) is not equal to zero.

Use definition of Residue

The definition of the residue at \(z_0\) for a function which can be expressed as a ratio of two functions \(g(z)\) and \(h(z)\) is given by \(Res[f(z); z_0] = lim_{z -> z_0} (z - z_0)f(z)\). Apply this definition by substituting \(f(z) = g(z)/h(z)\). We have \(Res[f(z); z_0] = lim_{z -> z_0} (z - z_0)\frac{g(z)}{h(z)}\). Since \(h(z_0) = 0\), this limit forms an indeterminate form.

L'Hopital's Rule

To resolve the indeterminate form, we can make use of L'Hopital's rule. Applying the rule, we get \(Res[f(z); z_0] = lim_{z -> z_0} \frac{g(z)}{h'(z)}\). Evaluating this limit at \(z_0\), we obtain \(Res[f(z); z_0] = \frac{g(z_0)}{h'(z_0)}\) which proves the theorem.

Key concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis that helps in evaluating complex integrals. It's particularly useful when dealing with integrals of functions that have poles, which are points where the function goes to infinity.

The key idea behind the residue theorem is that it relates the integral of a function around a closed contour to the sum of residues of the function inside the contour. A residue is essentially the coefficient of the \/(1\/(z - z_0)\/) term in the Laurent series expansion of a function around a singularity at \(z_0\).

In the original exercise, the residue at \(z_0\) is determined for the function \(f(z) = \frac{g(z)}{h(z)}\). The given formula \(\operatorname{Res}\left[\frac{g(z)}{h(z)}; z_{0}\right] = \frac{g(z_0)}{h'(z_0)}\) provides us with a direct way to compute this residue. This computation is crucial because residues simplify the evaluation of complex integrals that circle around poles.

Complex Differentiation

Complex differentiation is akin to differentiation in real analysis, but it involves complex-valued functions. A function \(f(z)\) is complex differentiable at a point if the limit defining the derivative exists. The function \(f(z)\) is then said to be analytic at that point.

Analytic functions possess several rich properties:

• Smoothness: They are infinitely differentiable.
• Conformality: They preserve angles and shapes (locally).
• Power Series Representation: Analytic functions can be expressed as power series within their radius of convergence.

In the context of the exercise, both \(g(z)\) and \(h(z)\) are analytic at \(z_0\). This means they have well-defined derivatives at that point, which helps in simplifying the computation of the residue. Analyticity ensures that \(g(z_0) eq 0\) and \(h'(z_0)eq 0\), allowing the use of the residue formula directly.

L'Hopital's Rule

L'Hopital's Rule is a mathematical tool used to find limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that if \( \lim_{x \to a} \frac{f(x)}{g(x)}\) results in these forms, then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]provided the limit on the right side exists.

In the problem, to evaluate \( \lim_{z \to z_0} (z - z_0) \frac{g(z)}{h(z)} \), L'Hopital's Rule resolves the \(\frac{0}{0}\) form resulting from \(h(z_0) = 0\). By differentiating the numerator and denominator, we transition to \(\lim_{z \to z_0} \frac{g(z)}{h'(z)}\).

Applying the rule simplifies the problem, resulting in the elegant expression \( \frac{g(z_0)}{h'(z_0)} \), which matches the residue as required by the problem statement. L'Hopital's Rule is essential here because it helps navigate through indeterminate forms to find a clear solution.