# What is the multiplicative inverse of 6-3i?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The multiplicative inverse of a term x is given by y such that x*y = 1.

Now we have to find the multiplicative inverse of 6 - 3i

Let it be x + yi

(6 - 3i)(x + yi) = 1

=> x + yi  = 1/(6 - 3i)

=> x + yi = (6 + 3i)/(6 - 3i)(6 + 3i)

=> x + yi = (6 + 3i)/ (6^2 + 3^2)

=> x + yi = (6+ 3i)( 36 + 9)

=> x + yi = (6 + 3i)/ 45

=> x + yi = 6/45 + (3/45)i

The multiplicative inverse is 6/45 + (3/45)i

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The multiplicative inverse is the fraction whose numerator is 1 and denominator is the given complex number.

We'll note the inverse as x:

(6-3i)*x = 1

We'll divide by (6-3i) both sides:

x = 1/(6-3i)

Since it is not allowed to keep a complex number to denominator, we'll multiply the ratio by the conjugate of the complex number:

x = (6+3i)/(6-3i)(6+3i)

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

(6-3i)(6+3i) = 6^2 - (3i)^2

(6-3i)(6+3i) = 36 - 9i^2

But i^2 = -1:

(6-3i)(6+3i) = 36 + 9

(6-3i)(6+3i) =  45

We'll re-write x:

x = (6+3i)/45

x = 6/45 + 3i/45

We'll simplify and we'll get:

x =  2/15 + i/15

The multiplicative inverse of the complex number 6 - 3i is 2/15 + i/15.