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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.

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  • $\begingroup$ You need to wrap the 50 in curly braces. $\endgroup$ – user856 Sep 5 '10 at 13:45
  • $\begingroup$ Is there a typo in $i^3+i^4+\dots +i^50$? Do you mean $i^3+i^4+\dots +i^{50}$? $\endgroup$ – Américo Tavares Sep 5 '10 at 13:47
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    $\begingroup$ Yes, sorry for the typo. $\endgroup$ – jkottnauer Sep 5 '10 at 13:52
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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, $i^3+i^4+\cdots+i^{50}=0$.

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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"

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  • $\begingroup$ Tavares: Its not an arithmetic progression but a geometric progression. $\endgroup$ – anonymous Sep 5 '10 at 13:57
  • $\begingroup$ @Chandru1: Of course! Thanks! Corrected. $\endgroup$ – Américo Tavares Sep 5 '10 at 14:10
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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: $\;\;\;\;\rm \displaystyle z^j \ (1+z+\cdots + z^{k-j}) \;=\; z^j \: \frac{1-z^{k-j+1}}{1-z} = 0 \;\iff\; \rm m\:|\:k-j+1\quad\;$

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    $\begingroup$ This is pretty opaque to a lowly peon such as myself! $\endgroup$ – The Chaz 2.0 Jul 16 '13 at 0:00
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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 $0$.

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