# How do you simplify (7i)/(8+i)?

Sep 2, 2016

#### Answer:

$\frac{7 i}{8 + i} = \frac{7}{65} + \frac{56}{65} i$

#### Explanation:

To simplify $\frac{7 i}{8 + i}$, we should multiply numerator and denominator by the complex conjugate of the denominator.

As complex conjugate of the denominator $\left(8 + i\right)$ is $\left(8 - i\right)$, we have

(7i)/(8+i)=(7i×(8-i))/((8+i)(8-i))

= $\frac{56 i - 7 {i}^{2}}{64 + 8 i - 8 i - {i}^{2}}$

= (56i-7×(-1))/(64+1)

= $\frac{7 + 56 i}{65}$

= $\frac{7}{65} + \frac{56}{65} i$

Sep 4, 2016

#### Answer:

$\frac{7}{65} + \frac{56}{65} i$

#### Explanation:

We have: $\frac{7 i}{8 + i}$

Let's multiply both the numerator and the denominator by the complex conjugate of the denominator:

$= \frac{7 i}{8 + i} \cdot \frac{8 - i}{8 - i}$

$= \frac{\left(7 i\right) \left(8\right) + \left(7 i\right) \left(- i\right)}{{\left(8\right)}^{2} - {\left(i\right)}^{2}}$

$= \frac{56 i - 7 {i}^{2}}{64 - {i}^{2}}$

Let's apply the fact that ${i}^{2} = - 1$:

$= \frac{56 i - \left(7 \cdot \left(- 1\right)\right)}{64 - \left(- 1\right)}$

$= \frac{56 i + 7}{65}$

$= \frac{7}{65} + \frac{56}{65} i$