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

Jan 28, 2016

Answer:

$\frac{5 - 8 i}{8 + 5 i} = - i$

Explanation:

Given:

$z = {z}_{1} / {z}_{2}$ with ${z}_{1} , {z}_{2} \in \mathbb{C}$

you can compute the Division of Complex Numbers multiplying both denominator an numerator by $\overline{{z}_{2}}$, the complex conjugate of ${z}_{2}$

If ${z}_{2} = a + i b \implies \overline{{z}_{2}} = a \textcolor{red}{-} i b$

$\therefore \frac{5 - 8 i}{8 + 5 i} = \frac{5 - 8 i}{8 + 5 i} \times \frac{8 - 5 i}{8 - 5 i} =$

$= \frac{5 \cdot 8 - 5 \cdot 5 i - 8 i \cdot 8 + 8 i \cdot 5 i}{{8}^{2} - {\left(5 i\right)}^{2}} = \frac{40 - 25 i - 64 i + 40 {i}^{2}}{64 - 25 {i}^{2}}$

remembering that ${i}^{2} = - 1$

$= \frac{\cancel{40} - \cancel{40} - 89 i}{64 + 25} = - \frac{\cancel{89} i}{\cancel{89}} = - i$