Compute $i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}$.
I notice that this is a geometric series with a common ratio of $i$, which is a complex number that satisfies $i^2=-1$. I also notice that the series has 201 terms, starting from $i^{-100}$ and ending at $i^{100}$. To find the sum of a geometric series, I can use the formula $S_n=a\frac{1-r^n}{1-r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. In this case, $a=i^{-100}$, $r=i$, and $n=201$. So the sum is $S_{201}=i^{-100}\frac{1-i^{201}}{1-i}$. To simplify this expression, I need to evaluate $i^{-100}$ and $i^{201}$. I recall that $i^0=1$, $i^1=i$, $i^2=-1$, and $i^3=-i$. This pattern repeats every four powers, so I can use modular arithmetic to find the remainder when dividing the exponent by 4. For example, $i^{-100}$ has a remainder of 0 when divided by 4, so it is equal to $i^0=1$. Similarly, $i^{201}$ has a remainder of 1 when divided by 4, so it is equal to $i^1=i$. Therefore, $S_{201}=1\frac{1-i}{1-i}$. This simplifies to $S_{201}=1$. 