How do you divide (4-8i)/(6-5i)?

Oct 16, 2016

=$\frac{64}{51} - \frac{28 i}{51}$
In this case the conjugate is $6 + 5 i$
$\frac{4 - 8 i}{6 - 5 i} = \frac{\left(4 - 8 i\right) \left(6 + 5 i\right)}{\left(6 - 5 i\right) \left(6 + 5 i\right)} = \frac{24 + 20 i - 48 i + 40}{51} = \frac{64 - 28 i}{51} = \frac{64}{51} - \frac{28 i}{51}$
$\left(6 + 5 i\right) \left(6 - 5 i\right) = 36 - 25 {i}^{2} = 36 + 25 = 51$ as ${i}^{2} = - 1$