Complex Power Series

**olski1** · August 22nd 2011, 11:41 PM

Question:

Let F(z) be the antiderivative of f(z)=e^(z^2)with F(0). Find F(z) as a power series around z=0.

Attempt at solution:

We know that, $e^z = \sum_{n=0}^{\infty} \frac{1}{n!}z^n$

Hence, $e^{z^2} = \sum_{n=0}^{\infty} \frac{1}{n!}z^{2n}$

Thus,

$F(z)=\int e^{z^2} = C+\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$

Now this is where i have problems, to find C,

let $e^{z^2}= (cos\theta +isin\theta)^2$

Integrating we find,

$\int(cos\theta +isin\theta)^2 = \frac{-1}{2}icos^2\theta+\frac{1}{2} sin2\theta=F(z)$

Therefore,

$F(z)= \frac{-1}{2}icos^2\theta+\frac{1}{2} sin2\theta=C+\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$

$=> F(0)=\frac{-i}{2}= C$

Therefore,

$F(z)= \frac{-i}{2} +\sum_{n=0}^{\infty} \frac{1}{(2n+1)n!}z^{2n+1}$

Is my methodology correct? there is no answer in the book for some reason, but upon research i think my power series is correct, but none of them have a value for C when doing an indefinate integral.

**girdav** · August 23rd 2011, 02:42 AM

Originally Posted by olski1

Question:

Let F(z) be the antiderivative of f(z)=e^(z^2)with F(0).

What do we assume about $F(0)$ ?

**olski1** · August 23rd 2011, 04:10 AM

Thats another thing i am unsure about. That is exactly the question given. If its a typo are you able to infer what they really meant to say, because i am unsure.

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