Question

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

Answer 1

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: `(2+i)/(2-i)`

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

`(2+i)/(2-i)(2+i)/(2+i)`

The denominator becomes the difference of two squares:

`((2+i)(2+i))/(4-i^2)`

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

`(4+ 2i+2i+i^2)/(4-i^2)`

Replace `i^2` with `-1`:

`(4+ 2i+2i-1)/(4-(-1))`

Combine like terms:

`(3+4i)/5`

Convert do `a+bi` form:

`3/5+4/5i`

Answer 2

`3/5+4/5i`

Explanation:

`color(orange)"Reminder " i^2=(sqrt(-1))^2=-1`

`"before dividing the complex numbers we require the"`
`"denominator to be real"`

`"this is achieved by multiplying the numerator/denominator"`
`"by the "color(blue)"complex conjugate "" of the denominator"`

`"Note " (a+bi)(color(red)(a-bi))=a^2+b^2larr" a real number"`

`"the conjugate has the same values of a and b with the"`
`"opposite sign"`

`"the conjugate of " (2-i)" is " (2+i)`

`rArr(2+i)/(2-i)xx((2+i))/((2+i))=((2+i)(2+i))/((2-i)(2+i))`

`"expanding numerator/denominator gives"`

`(4+4i+i^2)/(4-i^2)=(3+4i)/5`

`rArr(2+i)/(2-i)=3/5+4/5ilarrcolor(red)" in standard form"`