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 #50262 – Math – Complex Analysis

Given:

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

Find the taylor series of $f(z) = e^{\lambda} \left\{ z^{\lambda}2 + 2z - 3 \right\}$ about $z = -1$ along with its convergent neighborhood

Solution:

Method 1 (application of expansion for exponential function and product of series):

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

Method 2 (application of Taylor’s formula):

$$\begin{aligned} f(z) &= e^{z^2 + 2z - 3} = e^{(z + 1)^2 - 4} = \frac{1}{e^4} \cdot e^{(z + 1)^2} = \\ &= \frac{1}{e^4} \left( g(-1) + g'(-1)(z + 1) + \frac{g''(-1)}{2!}(z + 1)^2 + \dots + \frac{g^{(n)}(-1)}{n!}(z + 1)^n + \dots \right), \end{aligned}$$

where $g(z) = e^{(z + 1)^2}$, $g(-1) = e^0 = 1$,

$$\begin{aligned} g'(z) &= e^{(z + 1)^2}(2z + 2), \quad g'(-1) = e^0(-2 + 2) = 0, \\ g''(z) &= \left(e^{z^2 + 2z + 1}\right)(2z + 2) + e^{z^2 + 2z + 1}(2z + 2)' = e^{z^2 + 2z + 1}(2z + 2)^2 + 2e^{z^2 + 2z + 1}, \\ g''(-1) &= e^{z^2 + 2z + 1}(-2 + 2)^2 + 2e^{z^2 + 2z + 1} = 2, \\ g''(z) &= \left(e^{z^2 + 2z + 1}\right)(4z^2 + 8z + 6) + e^{z^2 + 2z + 1}(4z^2 + 8z + 6)' = \\ &= e^{z^2 + 2z + 1}(4z^2 + 8z + 6)(2z + 2) + e^{z^2 + 2z + 1}(8z + 8) \\ g''(-1) &= e^0(4 - 8 + 6)(-2 + 2) + e^0(-8 + 8) = 0 \end{aligned}$$

and so on.

$$f(z) = e^{z^2 + 2z - 3} = \frac{1}{e^4} \left(1 + (z + 1)^2 + \dots\right)$$

Method 3 (application of expansion for exponential function and multinomial formula):

Since $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$, then

$$\begin{array}{l} f (z) = e ^ {z ^ {2} + 2 z - 3} = e ^ {(z + 1) ^ {2} - 4} = \frac {1}{e ^ {4}} \cdot e ^ {(z + 1) ^ {2}} = \\ \frac {1}{e ^ {4}} \sum_ {n = 0} ^ {\infty} \frac {\left(\left(z + 1\right) ^ {2}\right) ^ {n}}{n !} = \frac {1}{e ^ {4}} \sum_ {n = 0} ^ {\infty} \frac {\left(z ^ {2} + 2 z + 1\right) ^ {n}}{n !} = \\ \sum_ {n = 0} ^ {\infty} \left(\frac {\sum_ {k + l + m = n , k \geq 0 , l \geq 0 , m \geq 0} \frac {n !}{k ! l ! m !} 2 ^ {l} z ^ {2 k} z ^ {l}}{n !}\right). \end{array}$$

Answer: $f(z) = e^{z^2 + 2z - 3} = \frac{1}{e^4}\left(1 + (z + 1)^2 + \ldots\right)$

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