Find the multiplicative inverse of the complex number 4 - 2i .Find the multiplicative inverse of the complex number 4 - 2i .

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to remember that you only may evaluate the multiplicative inverse of a complex number in its rectangular form, such that:

Multiplicative inverse of `a - b*i` is `1/(a -b*i)`

Reasoning by analogy yields:

Multiplicative inverse of `4 - 2*i` is `z' = 1/(4 - 2i).`

You need to perform the following multiplication, such that:

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

Converting the product `((4 + 2i)(4 - 2i))` into a difference of squares, yields:

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

Using complex number theory you need to substitute `-1` for `i^2` , such that:

`1/(4 - 2i) = (4+2i)/(16 + 4)`

`1/(4 - 2i) = (4+2i)/20 => 1/(4 - 2i) = (2(2+i))/20`

Reducing duplicate factors yields:

`1/(4 - 2i) = (2+i)/10`

Hence, evaluating the multiplicative inverse of the given complex number `4 - 2i` yields `(2+i)/10.`

giorgiana1976 | Student

The multiplicative inverse is get when we multiply the given complex number by the inverse number we'll get the result 1.

We'll note the inverse as x:

(4-2i)*x = 1

We'll divide by (4-2i) both sides:

x = 1/(4-2i)

We are not allowed to keep a complex number to denominator, so we'll multiply the ratio by the conjugate of the complex number:

x = (4+2i)/(4-2i)(4+2i)

We'll re-write the denominator as a difference of squares:

(4-2i)(4+2i) = 4^2 - (2i)^2

(4-2i)(4+2i) = 16 - 4i^2

But i^2 = -1:

(4-2i)(4+2i) = 16 + 4

(4-2i)(4+2i) = 20

We'll re-write x:

x = (4+2i)/20

x = 4/20 + 2i/20

We'll simplify and we'll get:

x =  1/5 + i/10

The multiplicative inverse of the complex number 4 - 2i is 1/5 + i/10.