We are given that $A = 1 + i + i^2 + ... + i^{2009}$. We are also given that $B = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $C = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. We want to find the value of $(B^2 - C + i)^{20}$.

1. Problem Description

We are given that $A = 1 + i + i^2 + ... + i^{2009}$. We are also given that $B = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $C = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. We want to find the value of $(B^2 - C + i)^{20}$.

2. Solution Steps

First, let's find $B^2$.

$B^2 = (-\frac{1}{2} + i\frac{\sqrt{3}}{2})^2 = (-\frac{1}{2})^2 + 2(-\frac{1}{2})(i\frac{\sqrt{3}}{2}) + (i\frac{\sqrt{3}}{2})^2 = \frac{1}{4} - i\frac{\sqrt{3}}{2} - \frac{3}{4} = -\frac{2}{4} - i\frac{\sqrt{3}}{2} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.

Notice that $B^2 = C$.

So, $B^2 - C + i = C - C + i = i$.

Then we want to find $(B^2 - C + i)^{20} = (i)^{20}$.

Recall that $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. Thus, the powers of $i$ cycle every 4 powers.

Since $20 = 4 * 5$, $i^{20} = (i^4)^5 = 1^5 = 1$.

3. Final Answer

$(B^2 - C + i)^{20} = 1$