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Resoudre l'exercice etape par etape et faire tous xalcules demandé

The exercise involves analyzing the series given by:

\[
f(z) = \sum_{n=0}^{\infty} \frac{z^{n+1}}{2^n}
\]

Let's go through each part of the exercise step-by-step.

---

### Part (a): **Déterminer le rayon de convergence de la série entière \( f(z) \)**

To determine the radius of convergence \( R \) of a power series, we can use the **formula**:

\[
\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{\left| a_n \right|}
\]

For this series, the general term is:
\[
a_n = \frac{1}{2^n}
\]

Thus, the series can be rewritten as:
\[
f(z) = \sum_{n=0}^{\infty} \left(\frac{z^{n+1}}{2^n}\right)
\]

Alternatively, the **ratio test** can also help determine the radius of convergence:
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]

Applying the ratio test for \( \frac{1}{2^n} \), we observe that it converges for all \( z \) values such that:
\[
\left| \frac{z}{2} \right| < 1 \Rightarrow |z| < 2
\]

Therefore, the **radius of convergence \( R \) is 2**.

---

### Part (b): **Quelle est la valeur du prolongement analytique de \( f \) au point \( z = 2 - 3i \)?**

Since the radius of convergence \( R = 2 \), the point \( z = 2 - 3i \) lies outside this radius (\( |2 - 3i| = \sqrt{2^2 + (-3)^2} = \sqrt{13} \approx 3.61 \), which is greater than 2).

This means that the **series does not converge** at \( z = 2 - 3i \), so \( f(z) \) cannot be analytically continued to that point using this series.

---

### Part (c): **Représenter ce prolongement analytique par une série de Taylor convergente au point \( z = 2 - 3i \)**

To represent the analytic continuation of \( f(z) \) around \( z = 2 - 3i \), we would typically need to find another series expansion (such as a Laurent series) around this point, as the original Taylor series centered at \( z = 0 \) does not converge beyond \( |z| = 2 \).

However, constructing such an expansion requires more context about the function \( f(z) \) beyond the initial series. Therefore, additional techniques or transformations (like Möbius transformation or Laurent series) would be needed for an explicit representation around \( z = 2 - 3i \).

---

### Part (d): **Que remarque-t-on ?**

The primary observation here is that:

1. The series \( f(z) \) has a limited radius of convergence (\( R = 2 \)), which restricts its convergence to points within this circle.
2. For points outside \( |z| = 2 \), such as \( z = 2 - 3i \), the series does not converge, highlighting the necessity of alternative methods (like analytic continuation) for evaluating \( f(z) \) beyond this boundary.

---

Would you like further explanations on any of these steps or have additional questions?

### Related Questions

1. How would the radius of convergence change if the series terms were modified?
2. What is the impact of complex points like \( z = 2 - 3i \) on convergence in general?
3. How does the ratio test help in determining the radius of convergence?
4. What is the difference between a Taylor series and a Laurent series for analytic continuation?
5. Can \( f(z) \) have different convergence behavior if centered at \( z = 2 - 3i \)?

### Tip

In problems involving series convergence, always start with the radius of convergence to understand the domain in which the series is valid before proceeding with specific evaluations.

Résoudre l'exercice étape par étape et faire tous les calculs demandés pour la série entière f(z) = ∑ (z^(n+1)) / (2^n).

Step-by-step solution for determining the radius of convergence, analytic continuation, and representation of a power series around a given complex point. This problem covers essential concepts in complex analysis, including power series convergence and analytic continuation.

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