How do you write 3/(4+i) in standard form?

Question

How do I rewrite this equation in a+bi format (standard form)? Looking for the steps involved.

Answer 1

Given a number such as   3/(4+i)

 

The complex conjugate of a + bi is a - bi

Given x/(a+bi) to eliminate the complex number in the denominator

we multiply by (a-bi)/(a-bi)

 

The complex conjugate of 4+i is 4-i

To get rid of the i in the denominator we multiply by

(4-i)/(4-i)

 

[3/(4+i)]/[(4-i)/(4-i)]

 

= [3(4-i)]/[(4+i)(4-i)]

 

= (12 - 3i)/[4(4)+4(-i)+4i-i²]

 

= (12-3i)/(16-i²)

 

= (12-3i)/(16+1)

 

= (12-3i)/17

 

= 12/17 - (3/17)i

Answer 2

times the top and bottom by (4-i)