I am reading introductory to complex analysis in an engineering class.
Solution 1:
We could manually calculate the first 20 terms for the maclaurin series $ f(z) = \exp(z^2) $ by using the definition of the maclaurin series. This would take forever though.
Beginning of solution 2:
We know that $C_{20} = \frac{f^{(20)}(0)}{20!}$ which implies that $f^{(20)}(0) = C_{20} \cdot 20!$.
If we calculate $C_{20}$ we have solved the problem.
Since $f(z)$ analytical everywhere (I think?) there exists a maclaurin series with an infinite radius of convergence with the coefficients: \begin{equation}C_{n} = \frac{1}{2\pi i} \int_{C_p} \frac{f(s)}{s^{n+1}} ds \end{equation}
where $C_p$ is the closed circle $|s| = p $ inside the radius of convergence for $f$.
For $n=20$ we have
\begin{equation}C_{20} = \frac{1}{2\pi i} \int_{C_p} \frac{f(s)}{s^{21}} ds \end{equation}
and with $f(s) = \exp(s^2)$ we get
\begin{equation}C_{20} = \frac{1}{2\pi i} \int_{C_p} \frac{\exp(s^2)}{s^{21}} ds. \end{equation}
Since $C_p$ is a circle we could parametrize it but I think that would make the integral very messy, so I have not attempted that yet. Any help appreciated.
Before posting solutions please note that this problem is based on an introductory course for engineers.
Since $$ e^z=\sum_{n=0}^{\infty}\frac{1}{n!}z^n $$ we have $$ e^{z^2}=\sum_{n=0}^{\infty}\frac{1}{n!}z^{2n}= \sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k $$ Comparing coefficients, which is possible because the Taylor (or MacLaurin, if you prefer) series is unique, we get $$ \frac{f^{(20)}(0)}{20!}=\frac{1}{10!} $$