Question

find the taylor series of f(z)= e^ { z^2 + 2z -3} about z=-1 along with its convergent neighborhood

Answer 1

Answer on the Question #50233, Math, Complex Analysis

Given:

$$f(z) = e^{z^2 + 2z - 3} \quad z_0 = -1$$

Solution:

$$f(z) = \frac{e^{z^2} \cdot e^{2z}}{e^3} = \frac{1}{e^3} \cdot e^{z^2 - 1 + 1} \cdot e^{2z + 2 - 2} = \frac{1}{e^3} \cdot e \cdot e^{z^2 - 1} \cdot e^{-2} \cdot e^{2z + 2} = \frac{1}{e^4} \cdot \sum_{n=0}^{\infty} \frac{(z^2 - 1)^n}{n!} \cdot \sum_{n=0}^{\infty} \frac{(2z + 2)^n}{n!} = \\ = \frac{1}{e^4} \cdot \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{(z^2 - 1)^k}{k!} \cdot \frac{(2z + 2)^{n-k}}{(n-k)!} \right)$$

Answer: $\frac{1}{e^4} \cdot \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{(z^2 - 1)^k}{k!} \cdot \frac{(2z + 2)^{n-k}}{(n-k)!} \right)$

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