# Question #0ae21

Oct 12, 2016

$\frac{1 + i}{i \left(2 + 3 i\right)} = - \frac{1}{13} - \frac{5}{13} i$

#### Explanation:

Given a complex number $z = a + b i$, the complex conjugate of $z$, denoted $\overline{z}$, is given by $\overline{z} = a - b i$. A useful property of the complex conjugate is that $z \overline{z} = {a}^{2} + {b}^{2} \in \mathbb{R}$. We can use this to eliminate the complex value from the denominator of the expression.

$\frac{1 + i}{i \left(2 + 3 i\right)} = \frac{1 + i}{2 i + 3 {i}^{2}}$

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

$= \frac{\left(1 + i\right) \left(- 3 - 2 i\right)}{\left(- 3 + 2 i\right) \left(- 3 - 2 i\right)}$

$= \frac{- 3 - 3 i - 2 i - 2 {i}^{2}}{{\left(- 3\right)}^{2} + {2}^{2}}$

$= \frac{- 1 - 5 i}{13}$

$= - \frac{1}{13} - \frac{5}{13} i$