Calculate 25/(4+3i) +25/(4-3i)? Solution We\'ll factorize by 25 and we\'ll get: 25[1/(4+3i) + 1/(4-3i)] We\'ll multiply by (4+3i)(4-3i) = 4^2 - (3i)^2: 25[1/(4+3i) + 1/(4-3i)] = 25[(4-3i+4+3i)/(4+3i)*(4-3i)] We\'ll eliminate and combine like terms: 25[(4-3i+4+3i)/(4+3i)*(4-3i)] = 25*8/[4^2 - (3i)^2] But i^2 = -1: 25*8/[4^2 - (3i)^2] = 200/(16+9) 25/(4+3i) +25/(4-3i) = 200/25 25/(4+3i) +25/(4-3i) = 8 The result of addition of the complex numbers is the real number 8..