I need to find the power series for $e^z + e^{az} + e^{a²z}$ where $a$ is the complex number $e^{2πi/3}$.
I know that $1 + a + a² = 0$.
I have tried to differentiate the expression and give values to z but it doesn't get to anything satisfying. I have tried to write each term as a power series for the exponential function but, once I have this expression, I don't see how to deal with it.
Any help or hint would be much appreciated ! Thank you!
On
$$1+x+x^2=\frac{x^3-1}{x-1}$$ $$1+a^k+a^{2k}=\frac{a^{3k}-1}{a^k-1}$$
$$\text{Now since } a^{3k}=1 \text{ and } a^k\ne 1, \text{for all integers k not divisible by 3}$$
$$\text{ We get that the above expression is zero when $k$ is not divisible by $3$}$$
$$\text{ But when k is divisible by $3$ we have that } (1+a^k+a^{2k}=3)$$
$$\text{ So we have:}$$
$$\frac{1}{3}(1+a^k+a^{2k})= \begin{cases} 0 & \text{if $3\nmid k$}, \\ 1 & \text{if $3\mid k$ }. \end{cases} $$
$$e^{z}+e^{az}+e^{a^2z}=3\sum_{n=0}^\infty\frac{\frac{1}{3}(1+a^k+a^{2k})z^n}{n!}$$
So that we have: $$e^{z}+e^{az}+e^{a^2z}=3\sum_{n=0}^\infty \frac{z^{3n}}{(3n)!}$$
Using the fact that
$$e^{z}=\sum_{k=0}^{\infty}\frac{z^k}{k!},\ -\infty<|z|<\infty$$
Write
$$e^z+e^{az}+e^{a^2z}=\sum_{k=0}^{\infty}\frac{(1+a^{k}+a^{2k})z^k}{k!}$$
Can you carry on from here?
You can use your observation $1+a+a^2=0$ to simplify the sum further.