Question

Simplify 1/(6+5i) to get a complex number in standard a + bi form

Answer 1

multiply with conjugate

`(1)/(6 + 5i) * (6 - 5i)/(6 - 5i) = (6 - 5i)/(36 + 25) \\ = (6)/(61) - (5)/(61) i`

Answer 2

The required form is
`(6)/(61)-(5i)/(61)` or
`(6)/(61)+(-5i)/(61)`

Explanation:

Consider the provided complex number.

`(1)/((6+5i))`

Multiply the denominator and numerator with the conjugate of the denominator.

`(1)/((6+5i))* (6-5i)/(6-5i)`

`(6-5i)/((6)^2-(5i)^2)`

`(6-5i)/(36+25)`

`(6-5i)/(61)`

`(6)/(61)-(5i)/(61)` or
`(6)/(61)+(-5i)/(61)`

Hence, the required form is
`(6)/(61)-(5i)/(61)` or
`(6)/(61)+(-5i)/(61)`