<p>If \(p\) and \(q\) are distinct prime numbers, then the number of distinct imaginary numbers which are \(p\)th as well as \(q\)th roots of unity are</p>
Step-by-Step Solution
Key Concept: The pth roots of unity are e^(2πik/p) for k=0,1,...,p-1, and qth roots of unity are e^(2πij/q) for j=0,1,...,q-1. A number is both a pth and qth root of unity if and only if it is a gcd(p,q)th root of unity. Since p and q are distinct primes, gcd(p,q)=1, so the only common root is 1 (real), leaving gcd(p,q)-1=0 imaginary roots.
<p><strong>Step 1:</strong> The pth roots of unity are {e^(2πik/p) : k = 0, 1, 2, ..., p-1} and satisfy z^p = 1.</p><p><strong>Step 2:</strong> The qth roots of unity are {e^(2πij/q) : j = 0, 1, 2, ..., q-1} and satisfy z^q = 1.</p><p><strong>Step 3:</strong> A complex number is both a pth and qth root of unity if and only if it is a gcd(p,q)th root of unity (it must satisfy both z^p = 1 and z^q = 1, which means z^gcd(p,q) = 1).</p><p><strong>Step 4:</strong> Since p and q are distinct prime numbers, gcd(p,q) = 1. Therefore, the only number that is both a pth and qth root of unity is the 1st root of unity, which is z = 1 (a real number).</p><p><strong>Step 5:</strong> The number of distinct imaginary roots among the gcd(p,q)=1 roots of unity is 1 - 1 = 0 (we exclude the single real root 1).</p><p>∴ Answer: <strong>D (0)</strong></p>
Correct Answer: D