# Finding power series

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!

• Start with a series expansion of $e^z$. From that, you should be able to find a series expansion of $e^{az}$ and $e^{a^2 z}$. Finally, argue that the sum of expansions is the expansion of the sum, and use that to find the expansion of $e^z + e^{az} + e^{a^2 z}$. Finally, put in your special $a$ and simply...
– fgp
Commented Jan 7, 2014 at 12:47

$$1+x+x^2=\frac{x^3-1}{x-1}$$ $$1+a^k+a^{2k}=\frac{a^{3k}-1}{a^k-1}$$

$$\text{Now since } a^{3k}=1 \text{ and } a^k\ne 1, \text{for all integers k not divisible by 3}$$

$$\text{ We get that the above expression is zero when k is not divisible by 3}$$

$$\text{ But when k is divisible by 3 we have that } (1+a^k+a^{2k}=3)$$

$$\text{ So we have:}$$

$$\frac{1}{3}(1+a^k+a^{2k})= \begin{cases} 0 & \text{if 3\nmid k}, \\ 1 & \text{if 3\mid k }. \end{cases}$$

$$e^{z}+e^{az}+e^{a^2z}=3\sum_{n=0}^\infty\frac{\frac{1}{3}(1+a^k+a^{2k})z^n}{n!}$$

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.

• Hum.. I've got there but I don't know how to relate the sums 1 + a + a² and 1 + a^k + a^2k in order to simplify the sum...
– Kika
Commented Jan 7, 2014 at 13:13