<|system|>
For complex numbers $z$, let \[f(z) = \left\{
\begin{array}{cl}
z^{2}&\text{ if }z\text{ is not real}, \\
z+2 &\text{ if }z\text{ is real}.
\end{array}
\right.\]Find $f(i)+f(1)+f(-1)+f(-i)$.</s>.
<|user|>
Can you answer based on context?
<|assistant|>
I can use the fact that $i$ and $-i$ are imaginary numbers, and $1$ and $-1$ are real numbers, to apply the appropriate rule for each term. For $f(i)$, I use the rule for non-real numbers, which is $z^2$. So $f(i) = i^2 = -1$. Similarly, for $f(-i)$, I use the same rule, and get $f(-i) = (-i)^2 = -1$. For $f(1)$, I use the rule for real numbers, which is $z+2$. So $f(1) = 1+2 = 3$. For $f(-1)$, I use the same rule, and get $f(-1) = -1+2 = 1$. Now I have all the terms I need to add up: $f(i)+f(1)+f(-1)+f(-i) = -1+3+1-1 = 2$.

