# How do you divide (3i)/(1+i) + 2/(2+3i) ?

Jan 28, 2016

$\frac{27}{26} i + \frac{47}{26}$

#### Explanation:

$\frac{3 i}{1 + i} + \frac{2}{2 + 3 i}$
=(3i(1-i))/((1+i)(1-i))+(2(2-3i))/((2+3i)(2-3i)
$= \frac{3 i - 3 {i}^{2}}{1 - {i}^{2}} + \frac{4 - 6 i}{4 - 9 {i}^{2}}$
$= \frac{3 i - 3 \left(- 1\right)}{1 - \left(- 1\right)} + \frac{4 - 6 i}{4 - 9 \left(- 1\right)}$ [as, ${i}^{2} = - 1$ ]
$= \frac{3 i + 3}{1 + 1} + \frac{4 - 6 i}{4 + 9}$
$= \frac{3 i + 3}{2} + \frac{4 - 6 i}{13}$
$= \frac{3}{2} i + \frac{3}{2} + \frac{4}{13} - \frac{6}{13} i$
$= \frac{27}{26} i + \frac{47}{26}$

Jan 28, 2016

$\frac{47}{26} + \frac{27}{26} i$

#### Explanation:

Get a common denominator.

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

$= \frac{6 i + 9 {i}^{2}}{2 + 5 i + 3 {i}^{2}} + \frac{2 + 2 i}{2 + 5 i + 3 {i}^{2}}$

$= \frac{2 + 8 i + 9 {i}^{2}}{2 + 5 i + 3 {i}^{2}}$

This can be simplified since ${i}^{2} = - 1$.

$= \frac{2 + 8 i - 9}{2 + 5 i - 3}$

$= \frac{- 7 + 8 i}{- 1 + 5 i}$

Now, multiply by the complex conjugate of the denominator.

$= \frac{- 7 + 8 i}{- 1 + 5 i} \left(\frac{- 1 - 5 i}{- 1 - 5 i}\right)$

$= \frac{7 + 27 i - 40 {i}^{2}}{1 - 25 {i}^{2}}$

$= \frac{7 + 27 i + 40}{1 + 25}$

$= \frac{47 + 27 i}{26}$

$= \frac{47}{26} + \frac{27}{26} i$