idontknow Dec 2015 973 128 Earth Jan 4, 2020 #1 Compute Z in terms of N . \(\displaystyle z=i+i^2 +i^3 +\dotsc + i^N \; , \: N\in \mathbb{N}.\)

romsek Math Team Sep 2015 2,885 1,609 USA Jan 4, 2020 #2 there's perhaps a fancier formula but $z_N = \begin{cases}i &N \pmod {4}=1\\-1 + i &N \pmod{4}=2\\-1 &N \pmod{4} = 3\\0 &N\pmod{4}=0\end{cases}$ Reactions: idontknow and Maschke

there's perhaps a fancier formula but $z_N = \begin{cases}i &N \pmod {4}=1\\-1 + i &N \pmod{4}=2\\-1 &N \pmod{4} = 3\\0 &N\pmod{4}=0\end{cases}$

skipjack Forum Staff Dec 2006 21,387 2,410 Jan 5, 2020 #3 \(\displaystyle \frac{i^N - 1}{i + 1}\) Reactions: idontknow