# How do you divide (2+5i)/(5+2i)?

Jan 13, 2016

$\frac{20}{29} + \frac{21}{29} i$

#### Explanation:

To divide $\frac{2 + 5 i}{5 + 2 i}$, you need to find the complex conjugate of the denominator and extend the fraction with it.

This way, you will be able to "get rid" of the $i$ in the denominator.

Let me walk you through the process:

Your denominator is $5 + 2 i$, thus the complex conjugate is $5 - 2 i$.

To extend the fraction, you need to multiply both the numerator and the denominator with $5 - 2 i$:

$\frac{2 + 5 i}{5 + 2 i} = \frac{\left(2 + 5 i\right) \cdot \left(5 - 2 i\right)}{\left(5 + 2 i\right) \cdot \left(5 - 2 i\right)} = \frac{10 + 25 i - 4 i - 10 {i}^{2}}{{5}^{2} - {\left(2 i\right)}^{2}} = \frac{10 + 21 i - 10 {i}^{2}}{25 - 4 {i}^{2}}$

...remember that ${i}^{2} = - 1$ ...

$= \frac{10 + 21 i + 10}{25 + 4} = \frac{20 + 21 i}{29} = \frac{20}{29} + \frac{21}{29} i$

Jan 13, 2016

$\frac{20}{29} + \frac{21}{29} i$

#### Explanation:

Multiply the numerator and denominator by the complex conjugate of 5 + 2i . This ensures that the denominator is real.

The complex conjugate of a complex number a + bi is a - bi.

note that (a + bi )(a - bi ) = ${a}^{2} + a b i - a b i - b {i}^{2} = {a}^{2} + {b}^{2}$ which is real.

$\left(i = \sqrt{-} 1 \Rightarrow {i}^{2} = {\left(\sqrt{-} 1\right)}^{2} = - 1\right)$

rArr ((2 +5i )(5 - 2i))/((5 + 2i)(5 - 2i)

$= \frac{10 + 25 i - 4 i - 10 {i}^{2}}{25 - 4 {i}^{2}} = \frac{10 + 21 i + 10}{29}$

$= \frac{20 + 21 i}{29}$

$\Rightarrow \frac{2 + 5 i}{5 + 2 i} = \frac{20}{29} + \frac{21}{29} i$