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

##### 2 Answers

#### Explanation:

#"we require to "color(blue)"rationalise 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"#

Multiply by the complex conjugate of the denominator to get

#### 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

Multiply the numerators and multiply the denominators to get

Since

When you add and subtract numbers with imaginary parts, just treat

...and you're done!