This is a series problem where the terms are complex numbers. I am looking for a better approach to solving this problem.
If $\displaystyle z = \frac{1+i}{\sqrt2}$, Evaluate $1 + z + z^2 + ... + z^{20}$
The way I solved this was to evaluate the terms upto the 8th term. like below,
$$ \begin{align} z^2 &= i \\ z^3 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^4 &= -1 \\ z^5 &= -\frac{1}{\sqrt2} -\frac{1}{\sqrt2}i \\ z^6 &= -i \\ z^7 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^8 &= 1 \\ \end{align} $$
Then evaluating the first 8 terms, $$ S_7 = 1 + z + ... + z^7 = 0 $$
Which implies that, $$ S_20 = 0 + 0 + z^{17} + z^{18} + z^{19} + z^{20} = 1 + z + z^2 + z^3 $$
Thus giving the solution, $$S_20 = 1 + (1 + \sqrt2)i$$
I have a couple of questions about this.
Just looking at the series itself($1 + z + z^2 + ...$) makes me think that the series is similar to a binomial series. The way the terms cancelled out made me think of a telescopic series. Can you guys shed some light on this?
While this solution works it took me a while to get there. Is there a more elegant/less tedious way of doing this?
2 Answers
$\begingroup$- If $z \ne 1$, we have
$$1 + z + \ldots z^{n-1} = \frac{z^n-1}{z-1}$$
So it greatly reduces the number of computation
- Note that in your case $z = e^{\frac{i \pi}{4}}$, which makes it faster to compute the powers of $z$.
Applying those two ideas gets you the result.
$\endgroup$ 1 $\begingroup$Use the formula for the sum of a geometric series and the fact that $$ \bigl(\cos\theta + i\sin\theta)^{n} = \cos{n\theta} + i \sin{n \theta}$$
$\endgroup$ 1
