Question

Simplify 1/4-7i to get a complex number in standard a + bi form.

Answer 1

`(1)/(4-7i)=(4)/(65)+(7)/(65)i`

Explanation:

we are given

`(1)/(4-7i)`

Firstly, we will get rid of imaginary term from denominator

so, we will multiply conjugate to both top and bottom term

`(1)/(4-7i)=(1* (4+7i))/((4-7i)* (4+7i))`

`(1)/(4-7i)=(4+7i)/(4^2-(7i)^2)`

`(1)/(4-7i)=(4+7i)/(16+49)`

`(1)/(4-7i)=(4+7i)/(65)`

we can also write as

so, we get

`(1)/(4-7i)=(4)/(65)+(7)/(65)i`