Question

What is the number of ordered pairs of real numbers (a, b) such that \[ (a + bi)^{2002} = a - bi \]

• A. 1001
• B. 1002
• C. 2004
• D. 2002

Answer 1

The Correct Option is C

Let \(z = a + bi\), then \(\bar{z} = a - bi\) Given \(z^{2002} = \bar{z} \Rightarrow |z|^{2002} = |\bar{z}|\), Write \(z = re^{i\theta} \Rightarrow z^{2002} = r^{2002} e^{i2002\theta} = r e^{-i\theta}\) Equating: \[ r^{2001} e^{i2003\theta} = 1 \Rightarrow r^{2001} = 1, e^{i2003\theta} = 1 \Rightarrow \theta = \frac{2\pi k}{2003} \] So total = 2003 values of \(\theta\) with one extra when \(b = 0\), total = 2004 real (a, b) pairs.