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

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,945 questions. We can answer yours, too.

Ask a question