Question

Prove that the following complex number is purely real (4+3i/3+2i)(4-3i/3-2i)

Answer 1

Answer:

[(4+3i)/(3+2i)][(4-3i)/(3-2i)]

or, [(4+3i)(4-3i)]/[(3+2i)(3-2i)]

or, [16-9i^2]/[9-4i^2] [Since, (a + b)(a-b) = a^2-b^2]

or, (16 +9)/(9+4) [ i^2 = -1]

or, 25/13

since, the complex number contain no imaginary Part, it is proved that the complex number is purely real.