Adding powers of $i$ 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.
 A: 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: $\;\;\;\;\rm \displaystyle z^j \  (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$
A: Observing that $i^{3}+i^{4}+\ldots +i^{50}$ is a geometric progression with ratio $i$, first term $i^3$ and $50-3+1=48$ terms, we have
$i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}=i^{3}\times 
\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}$
$=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0$
Edit: "arithmetic" corrected to "geometric"
A: From $i^2=-1$ you get $i^{4n}=1$, $\quad i^{4n+1}=i$, $\quad i^{4n+2}=-1$ and $i^{4n+3}=-i$.
Then you just count your positive and negative multiples of $1$ and $i$.
In particular, $i^3+i^4+\cdots+i^{50}=0$.
A: We have $$i^{3} + i^{4} + i^{5} = 1 = i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} = i^{3} + i^{4} + \cdots +i^{4n+1}$$
Now $i^{50}=1 \times -1$, therefore we have the sum is $0$.
