Question

How do you simplify `\frac { 4- 3i } { 5+ 2i }` assuming that `i` is an imaginary number?

Answer 1

`14/29-23/29i`

Explanation:

`"we require to "color(blue)"rationalise the denominator"`

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

`"multiply numerator/denominator by "(5-2i)`

`•color(white)(x)i^2=(sqrt(-1))^2=-1`

`rArr(4-3i)/(5+2i)xx(5-2i)/(5-2i)`

`"expand the factors using FOIL"`

`=(20-23i+6i^2)/(25-4i^2)`

`=(14-23i)/29`

`=14/29-23/29ilarrcolor(red)" in standard form"`

Answer 2

Multiply by the complex conjugate of the denominator to get `(14-23i)/(29)`

Explanation:

We use "conjugates" to get messy things like square roots and imaginary numbers out of expressions. A conjugate is the same two-term expression with a different sign in the middle. Generally, we like to get those "messy terms" out of the denominators of fractions. Use the complex conjugate of `5+2i`, which is `5-2i`. Set it up like this:

`((4-3i)/(5+2i))*((5-2i)/(5-2i))`

Multiply the numerators and multiply the denominators to get

`((4-3i)(5-2i))/((5+2i)(5-2i))=(20-15i-8i+6i^2)/(25-4i^2)= (20-23i-6)/(25-4i^2)`
Since `i^2=-1, -4i^2=-4(-1)=4`

When you add and subtract numbers with imaginary parts, just treat `i` like a regular variable:

`(20-23i-6)/(25-4i^2)=(14-23i)/29`

...and you're done!