Question

Let \(f\) be analytic at \(z_{0}\) and \(f\left(z_{0}\right)=f^{\prime}\left(z_{0}\right)=\cdots=f^{(k)}\left(z_{0}\right)=0\). Show that the following function is analytic at \(z_{0}\) : $$g(z)=\left\\{\begin{array}{ll} \frac{f(z)}{\left(z-z_{0}\right)^{k+1}} & \text { for } z \neq z_{0} \\ \frac{f^{(k+1)}\left(z_{0}\right)}{(k+1) !} & \text { for } z=z_{0} \end{array}\right.$$

Short answer

The function \(g(z)\) has been shown to be differentiable, and hence is analytic, at \(z_0\).

Step-by-step solution

Writing out the Taylor series expansion

The first step is to write down the Taylor series expansion of the function \(f(z)\) at the point \(z_0\) which is given as:\[f(z) = f\left(z_{0}\right) + f^{\prime}\left(z_{0}\right)(z - z_{0}) + \cdots + \frac{f^{(k)}\left(z_{0}\right)}{k!}(z - z_{0})^k + \frac{f^{(k+1)}\left(z_{0}\right)}{(k+1)!}(z - z_{0})^{k+1} + \cdots\]

Substituting known values into the Taylor series

From the problem, it's given that \(f\left(z_{0}\right)=f^{\prime}\left(z_{0}\right)=\cdots=f^{(k)}\left(z_{0}\right)=0\). So the Taylor series expansion of \(f(z)\) simplifies to:\[f(z) = \frac{f^{(k+1)}\left(z_{0}\right)}{(k+1)!}(z - z_{0})^{k+1} + \cdots\]

Defining function \(g(z)\) in terms of \(f(z)\)

From the problem, we are also given that \(g(z) = \frac{f(z)}{\left(z-z_{0}\right)^{k+1}}\) for \(z \neq z_{0}\) and\(g(z) = \frac{f^{(k+1)}\left(z_{0}\right)}{(k+1) !}\) for \(z = z_{0}\). If we plug in the simplified Taylor series expansion of \(f(z)\) from step 2, we can see that \(g(z)\) is well-defined and continuous for all \(z\).

Showing analyticity at \(z_0\)

For a function to be analytic at a point, it must be differentiable in a neighbourhood of that point, including at the point itself. It's clear that \(g(z)\) is differentiable for \(z \neq z_0\) by the quotient rule. When \(z = z_0\), \(g(z)\) reduces to a constant, \(\frac{f^{(k+1)}\left(z_{0}\right)}{(k+1) !}\), which is also differentiable. So, \(g(z)\) is differentiable, and hence analytic, at \(z_0\).

Key concepts

Taylor Series Expansion

When you encounter the term _Taylor Series Expansion_, think of it as a powerful tool used to express a function as an infinite sum of terms, calculated from its derivatives at a specific point. For complex functions, this series gives a way to approach the function as if you're zooming in on its behavior near a particular point. The general formula for the Taylor series of a function, say at point \( z_0 \), is given by:

\[ f(z) = f\left(z_{0}\right) + f^{\prime}\left(z_{0}\right)(z - z_{0}) + \frac{f^{\prime\prime}\left(z_{0}\right)}{2!}(z - z_{0})^2 + \cdots \]

In the context of the given problem, the function \( f(z) \) simplifies greatly because many of its initial derivatives at \( z_0 \) are zero. This makes the Taylor series start at a higher power of \( (z - z_{0}) \). It's like the series skips over the lower power terms because their coefficients are zero. This simplification often makes calculations significantly easier, especially in proofs and derivations related to analyticity.

Analytic Function

An _analytic function_ is one that can be represented by a convergent power series in some neighborhood of each point in its domain. This implies that not only is the function smooth and continuous, but it also has derivatives of all orders. This property is quite strong and is crucial in complex analysis.

For our exercise, we say \( f \) is analytic at \( z_0 \) because we were given that its Taylor series is valid around this point. This analyticity guarantees that the function behaves nicely, allowing for its transformation into \( g(z) \) as described in the exercise. Essentially, \( g(z) \) respects the analytic properties because it is derived from a transformation of another analytic function \( f(z) \).

• The ability to express \( g(z) \) both as a transformed series and a simplified constant at \( z_0 \) shows that the analyticity of \( f \) is preserved in \( g \).
• Analyticity involves differentiability and continuity at all considered points and even beyond.

Overall, understanding the condition of analyticity helps identify pathways to solving more complex function behavior questions.

Differentiability

In complex analysis, _differentiability_ means more than just having a derivative at a point. It requires the existence of derivatives in a neighborhood, leading towards the property of being analytic. A function is _differentiable_ in the complex sense if it satisfies the Cauchy-Riemann equations, indicative of how the function changes in infinitesimally small regions.

To determine that \( g(z) \) is analytic at \( z_0 \), it is necessary to verify that it is differentiable. From the solution, for points where \( z eq z_0 \), \( g(z) \) is expressed as a quotient, which is straightforward to differentiate using quotient rule techniques:

1. **For \( z eq z_0 \)**: The continuity and differentiability follow from the rearrangement of \( f(z) \) over its Taylor expansion terms.2. **At \( z = z_0 \)**: \( g(z) \) simplifies to a constant value, where differentiability holds naturally because constants have derivatives of zero, thus satisfying differentiability trivially at this specific point.

Therefore, knowing the differentiability at each point around \( z_0 \) provides solid evidence that the function \( g(z) \) is analytically valid at this singular point and throughout its defined region, fulfilling the requirements of differentiability in complex domains.