To solve the problem of finding the number of ordered pairs \((a, b)\) of real numbers such that \((a + ib)^{2008} = a - ib\), we can follow these steps:
### Step 1: Define the complex number
Let \( z = a + ib \). Then, the equation can be rewritten as:
\[
z^{2008} = \overline{z}
\]
where \(\overline{z} = a - ib\) is the complex conjugate of \(z\).
### Step 2: Take the modulus of both sides
Taking the modulus of both sides gives:
\[
|z|^{2008} = |\overline{z}|
\]
Since the modulus of a complex number and its conjugate are equal, we have:
\[
|z|^{2008} = |z|
\]
### Step 3: Analyze the modulus equation
This leads us to the equation:
\[
|z|^{2008} - |z| = 0
\]
Factoring out \(|z|\), we get:
\[
|z| (|z|^{2007} - 1) = 0
\]
### Step 4: Solve for \(|z|\)
From this equation, we have two cases:
1. \(|z| = 0\)
2. \(|z|^{2007} - 1 = 0\) which simplifies to \(|z| = 1\)
### Step 5: Case 1: \(|z| = 0\)
If \(|z| = 0\), then:
\[
z = 0 + i \cdot 0 \quad \Rightarrow \quad (a, b) = (0, 0)
\]
This gives us one ordered pair: \((0, 0)\).
### Step 6: Case 2: \(|z| = 1\)
If \(|z| = 1\), then we can express \(z\) in polar form:
\[
z = e^{i\theta} \quad \text{for some } \theta
\]
Thus, we have:
\[
z^{2008} = e^{i2008\theta} = \overline{z} = e^{-i\theta}
\]
This implies:
\[
e^{i2008\theta} = e^{-i\theta}
\]
### Step 7: Equate the arguments
From this, we can equate the arguments:
\[
2008\theta = -\theta + 2k\pi \quad \text{for some integer } k
\]
This simplifies to:
\[
2009\theta = 2k\pi \quad \Rightarrow \quad \theta = \frac{2k\pi}{2009}
\]
### Step 8: Count the distinct angles
The values of \(k\) can range from \(0\) to \(2008\) (inclusive), giving us \(2009\) distinct angles.
### Step 9: Find the corresponding pairs \((a, b)\)
For each angle \(\theta\), we can find the corresponding \(a\) and \(b\):
\[
a = \cos\left(\frac{2k\pi}{2009}\right), \quad b = \sin\left(\frac{2k\pi}{2009}\right)
\]
Thus, for each \(k\), we have a unique ordered pair \((a, b)\).
### Step 10: Total ordered pairs
Combining both cases, we have:
- 1 ordered pair from \(|z| = 0\)
- 2009 ordered pairs from \(|z| = 1\)
Therefore, the total number of ordered pairs \((a, b)\) is:
\[
1 + 2009 = 2010
\]
### Final Answer
The number of ordered pairs \((a, b)\) of real numbers such that \((a + ib)^{2008} = a - ib\) is **2010**.
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