# How do you divide #(2+5i)/(5+2i)#?

##### 2 Answers

#### Explanation:

To divide

This way, you will be able to "get rid" of the

Let me walk you through the process:

Your denominator is

To extend the fraction, you need to multiply both the numerator and the denominator with

#(2+5i)/(5+2i) = ((2 + 5i) * (5-2i)) / ((5+2i) * (5 - 2i)) = (10 + 25 i - 4 i - 10 i ^2) / (5^2 - (2i)^2) = (10 + 21 i - 10 i^2) / (25 - 4i^2)#

...remember that

# = (10 + 21i + 10) / (25 + 4) = (20 + 21i) / 29 = 20/ 29 + 21/29 i#

#### Explanation:

Multiply the numerator and denominator by the complex conjugate of 5 + 2i . This ensures that the denominator is real.

The complex conjugate of a complex number a + bi is a - bi.

note that (a + bi )(a - bi ) =

#( i = sqrt- 1 rArri^2 = (sqrt- 1 )^2 = - 1 ) #

#=( 10 + 25i -4i - 10i^2)/(25 -4i^2) =( 10 + 21i + 10)/29 #

#=( 20 + 21i)/29 #

#rArr( 2 + 5i)/(5 + 2i ) = 20/29 + 21/29 i #