We are given the complex number $A = 1 + i + i^2 + \dots + i^{2009}$ and we are asked to determine if $A$ is a purely imaginary number.

1. Problem Description

We are given the complex number $A = 1 + i + i^2 + \dots + i^{2009}$ and we are asked to determine if $A$ is a purely imaginary number.

2. Solution Steps

The given expression is a geometric series with first term $a = 1$, common ratio $r = i$, and $n = 2010$ terms. The formula for the sum of a finite geometric series is:

$S_n = \frac{a(1 - r^n)}{1 - r}$

In our case, we have:

$A = \frac{1(1 - i^{2010})}{1 - i} = \frac{1 - i^{2010}}{1 - i}$

We know that $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. So the powers of $i$ repeat every 4 terms.

We can find $i^{2010}$ by dividing 2010 by 4: $2010 = 4(502) + 2$, so $i^{2010} = i^{4(502) + 2} = (i^4)^{502} \cdot i^2 = (1)^{502} \cdot (-1) = -1$.

Therefore, $A = \frac{1 - (-1)}{1 - i} = \frac{2}{1 - i}$.

To express this in the form $a + bi$, we multiply the numerator and denominator by the conjugate of the denominator, which is $1 + i$:

$A = \frac{2}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{2(1 + i)}{(1 - i)(1 + i)} = \frac{2(1 + i)}{1 - i^2} = \frac{2(1 + i)}{1 - (-1)} = \frac{2(1 + i)}{2} = 1 + i$.

A purely imaginary number is a complex number of the form $bi$, where $b$ is a real number. Since $A = 1 + i$, the real part is 1 and the imaginary part is

1. Therefore, $A$ is not a purely imaginary number.

3. Final Answer

$A$ is not a purely imaginary number.