How do you write the following quotient in standard form (2+i)/(2-i)?

May 21, 2017

My preferred way to divide complex numbers is to multiply numerator and denominator by the complex conjugate of the denominator and then simplify.

Explanation:

Given: $\frac{2 + i}{2 - i}$

Multiply numerator and denominator by the complex conjugate of the denominator:

$\frac{2 + i}{2 - i} \frac{2 + i}{2 + i}$

The denominator becomes the difference of two squares:

$\frac{\left(2 + i\right) \left(2 + i\right)}{4 - {i}^{2}}$

Use the F.O.I.L method to multiply the the numerator:

$\frac{4 + 2 i + 2 i + {i}^{2}}{4 - {i}^{2}}$

Replace ${i}^{2}$ with $- 1$:

$\frac{4 + 2 i + 2 i - 1}{4 - \left(- 1\right)}$

Combine like terms:

$\frac{3 + 4 i}{5}$

Convert do $a + b i$ form:

$\frac{3}{5} + \frac{4}{5} i$

May 21, 2017

$\frac{3}{5} + \frac{4}{5} i$

Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} {i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1$

$\text{before dividing the complex numbers we require the}$
$\text{denominator to be real}$

$\text{this is achieved by multiplying the numerator/denominator}$
$\text{by the "color(blue)"complex conjugate "" of the denominator}$

$\text{Note " (a+bi)(color(red)(a-bi))=a^2+b^2larr" a real number}$

$\text{the conjugate has the same values of a and b with the}$
$\text{opposite sign}$

$\text{the conjugate of " (2-i)" is } \left(2 + i\right)$

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

$\text{expanding numerator/denominator gives}$

$\frac{4 + 4 i + {i}^{2}}{4 - {i}^{2}} = \frac{3 + 4 i}{5}$

$\Rightarrow \frac{2 + i}{2 - i} = \frac{3}{5} + \frac{4}{5} i \leftarrow \textcolor{red}{\text{ in standard form}}$