Question

show that 1-2i/3-4i+1+2i/3+4i is real​

Answer 1

SOLUTION

Mate here is your Solution please mark braineliest

$\frac{1 - 2i}{3 - 4i} + \frac{1 + 2i}{3 + 4i} \\ = \frac{(1 - 2i)(3 + 4i) + (1 + 2i)(3 - 4i)}{ {3}^{2} - {(4i)}^{2} } \\ = \frac{3 + 4i - 6i - 8 {i}^{2} + 3 - 4i + 6i - 8 {i}^{2} }{9 - 4 \times ( - 1) } \\ = \frac{6 - 16 \times ( - 1) }{9 + 4} \\ = \frac{22}{13}$

Therefore , the given expression is real