Thread: Complex Power Series

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

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

Thus,

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

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

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

Integrating we find,

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

Therefore,

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

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

Therefore,

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

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.

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 $\displaystyle F(0)$?

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.