TITLE: Adding powers of $i$
QUESTION [8 upvotes]: I've been struggling with figuring out how to add powers of $i$.
For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...
It has to do something with division by 4, since the "power cycle" of $i$ repeats every fourth power.
Thank you for any clues.

REPLY [8 votes]: HINT $\rm\quad\quad i^3 + \: i^4 \; + \:\;\cdots\;\: + \; i^k = 0\ \:\iff\: k\:\equiv\: 2 \:\pmod 4$
Generally, suppose that $\rm\: z \:$ has order $\rm m>1\:.$ Therefore $\rm\; z^n = 1 \iff\ m\:|n\;\;\:$ hence:
LEMMA $\quad\rm z^j + z^{j+1} + \:\cdots + z^k = 0\;\; \iff \rm\: k \:\equiv\;\; j \:-\: 1 \:\pmod m $
Proof: $\;\;\;\;\rm \displaystyle z^j \  (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$