Show that $e^{2z^3}$ is entire. Then find its derivative.

Question

I am currently teaching extension mathematics to a class and was posed this question by a very bright student. The text we use to teach this level of mathematics is Cambridge Mathematics 3 Unit and it briefly touches on these types of questions, but it is not something that is generally taught. Looking for some assistance as I do not want to steer my student in the wrong direction!

How can I best go about proving that $e^{2z^3}$ is entire? Then find the derivative?

Answer 1

$$e^z=\sum_{n\geq 0}\frac{z^n}{n!}\tag{1}$$ $$e^{2z^3}=\sum_{n\geq 0}\frac{2^n z^{3n}}{n!}\tag{2}$$ $$ \frac{d}{dz}e^{2z^3} = \sum_{n\geq 1}\frac{2^n\cdot 3n z^{3n-1}}{n!} = 6z^2 e^{2z^3}\tag{3} $$ and the radius of convergence of all these power series is pretty clearly $+\infty$.

Answer 2

Note that $(f\circ g)'(z)=f'(g(z))g'(z)$. Now $e^{(\cdot)}$ is entire, $2(\cdot)^{3}$ is entire, then so is the composition and $(e^{2z^{3}})'=e^{2z^{3}}(6z^{2})$.