Find integers p and q such that \(\displaystyle (3 + 7i)(p + qi)\) is purely imaginary.
Attempt:
\(\displaystyle (3 + 7i)(p + qi)\)
\(\displaystyle =3p + 3qi + 7pi - 7q\)
\(\displaystyle =3p - 7q + 3qi + 7pi\)
Since 3p - 7q is the real part, it must = 0 therefore:
\(\displaystyle 3p - 7q = 0\)
\(\displaystyle p = \frac{7q}{3}\)
and
\(\displaystyle q = \frac{3p}{7}\)
but the correct answer was p = 7n and q = 3n, where nEZ. I am confused :?
Attempt:
\(\displaystyle (3 + 7i)(p + qi)\)
\(\displaystyle =3p + 3qi + 7pi - 7q\)
\(\displaystyle =3p - 7q + 3qi + 7pi\)
Since 3p - 7q is the real part, it must = 0 therefore:
\(\displaystyle 3p - 7q = 0\)
\(\displaystyle p = \frac{7q}{3}\)
and
\(\displaystyle q = \frac{3p}{7}\)
but the correct answer was p = 7n and q = 3n, where nEZ. I am confused :?