Writing $\frac{1}{2}+i$ as $1+i+\frac{i^2}{2}$ in Needhams Complex Analysis text This question is from the first chapter of Needham's "Visual Complex Analysis Text"

My question is in regards to the sequence they get in the solution below.  Getting the first term in the sequence to be $1$ and the second being $1+i$, follows naturally, and when I think about the third term in the sequence it's clearly the complex number $\frac{1}{2}+i$.  How did they get the equivalent expression $1+i+\frac{i^2}{2}$?
Solution


The author used the following equivalences:
$$\begin{align}\text{East} &: i^0 =i^4=i^{8\ }=...=1\\
\text{North} &: i^1 =i^5=i^{9\ }=...=i\\
\text{West} &: i^2 =i^6=i^{10}=...=-1\\
\text{South} &: i^3 =i^7=i^{11}=...=-i\end{align}$$
And in general $i^k = i^{k\ \  (\text{ mod } 4)}$.
It's worth mentioning also that multiplication by $i$ is a rotation by $90^o$ or $\frac{\pi}{2}$ radians. By multiplying the $n^{th}$ move by $i$ and dividing by $n+1$, you are following the presented algorithm.
