# Simplify (Ci)  (2 - 3i) / (1 + 2i) and (Cii) (3i^12-i^9)/(2i+1)?

Mar 14, 2017

$\frac{2 - 3 i}{1 + 2 i} = - \frac{4}{5} - \frac{7}{5} i$

$\frac{3 {i}^{12} - {i}^{9}}{2 i + 1} = \frac{1}{5} - \frac{7}{5} i$

#### Explanation:

We can remove a complex denominator by multiplying the quotient by its complex conjugate (and therefore we must also multiply the numerator), This woks because the product of a complex number with its conjugate is always real, and ia a similar way to rationalising the denominator of an irrational expression to remove a surd denominator.

Question Ci

Here the conjugate of the denominator is $1 - 2 i$ so we get:

$\frac{2 - 3 i}{1 + 2 i} = \frac{\left(2 - 3 i\right)}{\left(1 + 2 i\right)} \cdot \frac{\left(1 - 2 i\right)}{\left(1 - 2 i\right)}$
$\text{ } = \frac{2 - 3 i - 4 i + 6 {i}^{2}}{1 + 2 i - 2 i - 4 {i}^{2}}$
$\text{ } = \frac{2 - 7 i - 6}{1 + 4} \setminus \setminus \setminus \setminus \setminus \left(\because {i}^{2} = - 1\right)$
$\text{ } = \frac{- 4 - 7 i}{5}$
$\text{ } = - \frac{4}{5} - \frac{7}{5} i$

Question Cii
Here the conjugate of the denominator is also $1 - 2 i$ so we get:

$\frac{3 {i}^{12} - {i}^{9}}{2 i + 1} = \frac{3 - i}{1 + 2 i}$
$\text{ } = \frac{\left(3 - i\right)}{\left(1 + 2 i\right)} \cdot \frac{\left(1 - 2 i\right)}{\left(1 - 2 i\right)}$
$\text{ } = \frac{3 - i - 6 i + 2 {i}^{2}}{1 + 2 i - 2 i - 4 {i}^{2}}$
$\text{ } = \frac{3 - 7 i - 2}{1 + 4}$
$\text{ } = \frac{1 - 7 i}{5}$
$\text{ } = \frac{1}{5} - \frac{7}{5} i$