How do you simplify (2+5i)/(5+2i) and write the complex number in standard form?

Jan 8, 2017

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

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So we find:

$\left(5 - 2 i\right) \left(5 + 2 i\right) = {5}^{2} - {\left(2 i\right)}^{2} = 25 + 4 = 29$

This provides a way to rationalise the denominator: Multiply both numerator and denominator by the complex conjugate of the denominator thus:

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

$\textcolor{w h i t e}{\frac{2 + 5 i}{5 + 2 i}} = \frac{\left(5 - 2 i\right) \left(2 + 5 i\right)}{29}$

$\textcolor{w h i t e}{\frac{2 + 5 i}{5 + 2 i}} = \frac{{\overbrace{5 \left(2\right)}}^{\text{First"+overbrace(5(5i))^"Outside"+overbrace((-2i)(2))^"Inside"+overbrace((-2i)(5i))^"Last}}}{29}$

$\textcolor{w h i t e}{\frac{2 + 5 i}{5 + 2 i}} = \frac{1}{29} \left(10 + 25 i - 4 i + 10\right)$

$\textcolor{w h i t e}{\frac{2 + 5 i}{5 + 2 i}} = \frac{1}{29} \left(20 + 21 i\right)$

$\textcolor{w h i t e}{\frac{2 + 5 i}{5 + 2 i}} = \frac{20}{29} + \frac{21}{29} i$