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.
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"
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, $i^3+i^4+\cdots+i^{50}=0$.
Hint $\quad i^3 + i^4 +\cdots+ i^k = 0\iff k\equiv 2 \:\pmod{\! 4}$
Generally suppose $\,z\,$ has order $\,m>1,\,$ thus $\,z^n = 1\iff m\mid n,\,$ so
Lemma $\ \ f := z^j + z^{j+1} + \cdots + z^k = 0 \iff k \equiv j- 1\pmod{\! m},\ $ for $\,z\neq 0,1$
Proof $\,\ f = z^j \ (1+z+\cdots + z^{k-j}) \;=\; z^j\,\ \dfrac{1-z^{N}}{\!\!\!1-z} = 0\iff m\mid\underbrace{ N\!=\!k\!-\!j\!+\!1}_{\!\#\rm summands}$
Generally for $\:\!z\,$ in any ring we have:
Theorem $\ \ m\mid N \,\Rightarrow\, f = 0, \ $ if $\:\color{#0a0}{\!1\!-\!z\ \rm is\ cancellable}$.
Conversely $\,\ f = 0^{\phantom{|^|}}\!\Rightarrow\, m\mid N,\,$ if $\, \color{#c00}{z\,\ \rm is\ cancellable}\,$ (or $\,j = 0)$
Proof $\,\ $ As above we have: $\ (1\!-\!z)f = z^j\,(1\!-\!z^N)$.
So $\ m\mid N\,\Rightarrow\, z^N = 1\,\Rightarrow\, (\color{#0a0}{1\!-\!z})f= 0,\,$ $\color{#0a0}{\rm so}\,\ f = 0,\,$
and, conversely, we have: $\ f = 0\,\Rightarrow \color{#c00}z^j(1\!-\!z^N) = 0,\ \color{#c00}{\rm so}\,\ z^N = 1,\,$ so $\:m\mid N$.
Corollary $\ f = 0\iff m\mid N,\,$ if $\,z\neq \color{#c00}0,\color{#0a0}1\,$ in a domain (e.g. in any field such as $\Bbb C$ in OP, or $\,\Bbb Z_p =$ integers $\!\bmod p\,$ prime), where elements are $\rm\color{#c00}{cancell}\color{#0a0}{able}$ iff invertible iff nonzero.
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 $0$.